# Triple Integral Pdf

(5 8 5) 4 5 60 3 3 3 x x x dx x x 3 2 9 5 9 2 2 1 1 2 1026 22 1001 2. View Math267-Triple-integrals_wNotes. But the real difficulty with triple integrals is-- and I think you'll see that your calculus teacher will often do this-- when you're doing triple integrals, unless you have a very easy figure like this, the evaluation-- if you actually wanted to analytically evaluate a triple integral that has more complicated boundaries or more complicated. Katz familiar to calculus students. Evaluating triple integrals A triple integral is an integral of the form Z b a Z q(x) p(x) Z s(x,y) r(x,y) f(x,y,z) dzdydx The evaluation can be split into an "inner integral" (the integral with respect to z between limits. 1 (Iterated Integrals). Then, parallel to the axis of walk,. 4 (approximate answer, depends on what you estimated the values at the midpoints to be). P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. INTEGRAL REVIEW July 2012 Vol. If it's a 2D surface, use a double integral. 1) is the signed volume bounded by the graph z f x y over the region; that is, the volume of the part of the solid below the xy-planeis taken to be negative. Triple integral is defined and explained through solved examples. Solutions to Midterm 1 Problem 1. Figure 1 In order for the double integral to exist, it is sufficient that, for example, the region D be a closed (Jordan) measurable region and that the function f(x, y ) be continuous throughout D. The inner limit is the easiest. Integration by Parts 21 1. • M yz = RRR S xδ(x,y,z)dV is the moment about the yz-plane. Cylindrical Coordinates, page 1040 In the cylindricalcoordinatesystem(柱坐標系), a point P in three-dimensional space is represented by the ordered triple (r,θ,z), where r and θ are polar coordinates of. Set up a triple integral of a function f(x,y,z) over a ball of radius 3 centered at (0,0,0) in R3. (So think of a wall around the perimeter of the ﬂoor area R, reaching up. above z = −√4x2 +4y2. 1 (Iterated Integrals). as an iterated integral (i. For any given θ, the angle φ that M makes with the z-axis runs from φ = φmin to φ = φmax. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. patrickJMT 357,008 views. Then, parallel to the axis of walk,. Multiple Integrals 15. By symmetry, ¯x = 0 and ¯y = 0, so we only need ¯z. The sphere x2 +y2 +z2 = 4 is the same as ˆ= 2. Let's choose the order y-z-x. Multiple Integrals 1. 3 4 4 22 1 1 5 188 8 1. analysis of Legendre polynomials triple product integral. Line integrals Z C dr; Z C a ¢ dr; Z C a £ dr (1) ( is a scalar ﬂeld and a is a vector ﬂeld)We divide the path C joining the points A and B into N small line elements ¢rp, p = 1;:::;N. Suppose that E is a "Type 1" region between surfaces with equations z = h(x, y) and z = k(x, y) and has perpendicular projection D on the xy-plane. You may discuss the problems with other students. [Hint: The volume of a (solid) region DˆR3 is RRR D 1dxdydz] Solution: We integrate ﬁrst with respect to z, keeping (x;y) ﬁxed. ) We will turn triple integrals into (triple) iterated integrals. 7: Triple Integrals Outcome A: Evaluate a triple integral by iterated integration. 8 The Fubini’s Method to evaluate triple integrals is this: We ﬁrst walk on a line parallel to one of the x-axis, y-axis, or z-axis. An orientable surface, roughly speaking, is one with two distinct sides. Calculadora gratuita de. Set up but do not evaluate the triple integral for the mass M of the solid bounded by the surface z = 1 x2 y2 and the xy plane, if the density is f(x;y;z) = x+1. Triple integration exercises 1. The volume is given by the. 1) is the signed volume bounded by the graph z f x y over the region; that is, the volume of the part of the solid below the xy-planeis taken to be negative. The dV in each of the integrals can be any of the 6 permutations of dx, dy, and dz. Proposition 17. ) Chapter 8 described the same idea for solids of revolution. The triple integral of a continuous function over a general solid region. See unit IV lesson 2 for a review. A good example to think about. The double integral of a nonnegative function f(x;y) deﬂned on a region in the plane is associated with the volume of the region under the graph of f(x;y). •Triple Integrals can also be used to represent a volume, in the same way that a double integral can be used to represent an area. Boise State Math 275 (Ultman) Worksheet 3. by evaluating the iterated integral using any of the six possible orders (Theorem 14. Cylindrical Coordinates, page 1040 In the cylindricalcoordinatesystem(柱坐標系), a point P in three-dimensional space is represented by the ordered triple (r,θ,z), where r and θ are polar coordinates of. Lady (December 21, 1998) Consider the following set of formulas from high-school geometry and physics: Area = Width Length Area of a Rectangle Distance = Velocity Time Distance Traveled by a Moving Object Volume = Base Area Height Volume of a Cylinder Work = Force Displacement Work Done by a Constant Force. (Hindi) Complete Engineering Mathematics for GATE 43 lessons • 5 h 53 m. Homework 6 Solutions Jarrod Pickens 1. For a triple integral in a region D ˆR3, it can be evaluated by using an iterated. The following concepts may or may not be seen on the exam and there may be concepts on the exam which are not covered on this sheet. I Project your region E onto one of the xy-, yz-, or xz-planes, and. It will come as no surprise that we can also do triple integrals---integrals over a three-dimensional region. 2 MATH11007 NOTES 18: TRIPLE INTEGRALS, SPHERICAL COORDINATES. : 0^ 2 ` E ³³³ yd 3. TRIPLE INTEGRALS IN SPHERICAL & CYLINDRICAL COORDINATES Triple Integrals in every Coordinate System feature a unique infinitesimal volume element. It's difficult to explain. Hence, the triple integral is given by Note that we can change the order of integration of r and theta so the integral can also be expressed Evaluating the iterated integral, we have find that the mass of the object is 1024*pi. Triple Integral Spherical Coordinates Pdf Download >>> DOWNLOAD (Mirror #1). This is an example of a triple or volume integral. That means we need to nd a function smaller than 1+e x. CALCULUS III DOUBLE & TRIPLE INTEGRALS STEP-BY-STEP A Manual For Self-Study prepared by Antony Foster Department of Mathematics (oﬃce: NAC 6/273) The City College of The City University of New York 160 Convent Avenue At 138th Street New York, NY 10031 [email protected] (b) Reverse the order of integration to dydzdx. Just as for double integrals, a region over which a triple integral is being taken may have easier representation in another coordinate system, say in uvw-space, than in xyz-space. Find materials for this course in the pages linked along the left. For example, the face of T in the xy-plane is given by x;y 0 and 2x + 3y 6. iosrjournals. Double and Triple Integrals. Fubini's theorem for triple integrals states that the value of a triple integral of a continuous function f over a region E in R 3 is a triple iterated integral. 3 Evaluate the integral RRR T. In the triple integral , , 0 If ( , , ) = 1 then this triple integral is the same as , which is simply the volume under the surface represented by z(x,y). Single Integral - the domain is the integral I (a line). (5), if ,, then the triple. 6: Triple Integrals Thursday, April 2, 2015 3:37 PM Section 15. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. The double integral JSf(x, y)dy dx will now be reduced to single integrals in y and then x. A volume integral is a specific type of triple integral. Areas and Distances. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each \$\Delta x\times\Delta y\times\Delta z. Compute volume, mass, and center of mass using triple integrals. (1) is deﬂned as Z C a ¢ dr = lim N!1 XN p=1 a(xp;yp;zp) ¢ rpwhere it is assumed that all j¢rpj ! 0. 1 (Iterated Integrals). Then for some continuous function f, f: \\3 →, the triple integral (),, R ∫∫∫f xyzdVcan be expressed 6 different ways in Cartesian (rectangular) coordinates. For example, nd out ∫ 1 0 1 (x+1) p x dx >## define the integrated function >integrand <- function(x) {1/((x+1)*sqrt(x))}. EXAMPLE 4 Find a vector field whose divergence is the given F function. Double and triple integrals 5 At least in the case where f(x,y) ≥ 0 always, we can imagine the graph as a roof over a ﬂoor area R. Using Iterated Integrals to find area. Triple integration of sum of two functions is explained. In this section we practice ﬁnding the integral of such functions. Today's Goals Today's Goals 1 Be able to set up and evaluate triple integrals in cartesian coordinates. Riemann Sums , Integral Representation for lengths, Areas, Volumes and Surface areas in Cartesian and polar coordinates multiple integrals – double and triple integrals – change of order of integration- change of variable. Both of the limits diverge, so the integral diverges. Thenthedeﬁnite. That means we need to nd a function smaller than 1+e x. La integral triple de f sobre la caja B es ZZZ B f(x,y,z)dV = l´ım l,m,n→∞ Xl i=1 m j=1 Xn k=1 f(x∗ ijk,y ∗ ijk,z ∗ ijk)∆V si el l´ımite existe. R/ Nótese que la región de integración es la parte de la esfera de centro en el origen de coordenadas y radio 1 que está contenida en el primer octante, que se muestra en la siguiente figura:. ) Chapter 8 described the same idea for solids of revolution. Suppose that w= f(x,y,z) is a continuous function on the rectangular parallelipiped R: a≤ x≤ b, c≤ y≤ d, p≤ z≤ q. , 0 ≤ y ≤ 1. Triple Integrals Part 1: De–nition of the Triple Integral We can extend the concept of an integral into even higher dimensions. ANSWERS TO REVIEW PROBLEMS A. Spherical coordinates are pictured below: The volume of the \spherical wedge" pictured is approximately V = ˆ2 sin˚ ˆ ˚: The ˆ2. Set up the integral. Triple Integrals and Triple Iterated Integrals section 13. The solid below is enclosed by x= 0, x= 1, y= 0, z= 0, z= 1, and 2x+y+2z= 6. which is an integral of a function over a two-dimensional region. f Triple Integrals in Cylindrical and Spherical Coordinates We have already seen the advantage of changing to polar coordinates in some double integral problems. This is somewhat subtle in the physical interpretation but can be summarized as "generality". The infinite series forms of the two types of triple integrals can be obtained using binomial series and integration term by term theorem. For example, nd out ∫ 1 0 1 (x+1) p x dx >## define the integrated function >integrand <- function(x) {1/((x+1)*sqrt(x))}. (iii) Change the limits of the integral and include the "r" in the integral. Step 1: Add one to the exponent. Similarly, if f(x,y,z. At any point on an orientable surface, there exists two normal vectors, one pointing in the opposite direction of the other. Integral calculus that we are beginning to learn now is called integral calculus. 2 Sketch the solid whose volume is given by the integral R 1 0 R 1−x 0 R 2−2z 0 dydzdx. the triple integral of f over the solid and denote it by RRR S f(x,y,z)dV. (b) Reverse the order of integration to dydzdx. Use the Comparison Theorem to decide if the following integrals are convergent or divergent. Usually these integrals cannot be solved. via contour integration. (So think of a wall around the perimeter of the ﬂoor area R, reaching up. Just as with double integrals, the only trick is determining the limits on the iterated integrals. I'm trying to do the same with triple integrals using Simpson's rule in VBA. The solution is found in terms of a function which is determined by means of a Fredholm integral equation of the first kind. It will cover three major aspects of integral calculus: 1. To show this, let g and h be two functions having the same derivatives on an interval I. In this section we provide a quick discussion of one such system — polar coordinates — and then introduce and investigate their ramifications for double integrals. The analogy between single and. Show Step-by-step Solutions. Our first integral could equally well be ff(x, y)dx. 1 DOUBLE INTEGRALS OVER RECTANGLES TRANSPARENCIES AVAILABLE #48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS 1 2 -1 class Essential Material POINTS TO STRESS 1. TRIPLE INTEGRALS IN SPHERICAL & CYLINDRICAL COORDINATES Triple Integrals in every Coordinate System feature a unique infinitesimal volume element. The paper also summarizes the results of the survey questions given to the students in two of the courses followed by the authors own critique of the enhancement project. Thank you! Part A: Triple. In particular, if then we have. 3 0 2ˇ 0 2 1 (r+ z)rdrd dz Region from Diagram 2 3 0 2ˇ 0 2 0 5zrdrd dz Region from Diagram 2 3 0 ˇ=2 0 2 1. BYJU’S online triple integral calculator tool makes the calculation faster, and it displays the integrated value in a fraction of seconds. line integrals, we used the tangent vector to encapsulate the information needed for our small chunks of curve. One should go to the original paper to admire the ingenuity displayed in ﬁnding (1. For example, nd out ∫ 1 0 1 (x+1) p x dx >## define the integrated function >integrand <- function(x) {1/((x+1)*sqrt(x))}. And a conversion can be made between 3 forms. We used a double integral to integrate over a two-dimensional region and so it shouldn’t be too surprising that we’ll use a triple integral to integrate over a three dimensional region. If we want to convert this triple integral to cylindrical coordinates we need to rewrite xand yusing the conversion formulas from above. projection of a function on i th and j th coordinates is calculated. Section 15. 1 DOUBLE INTEGRALS OVER RECTANGLES TRANSPARENCIES AVAILABLE #48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS 1 2 -1 class Essential Material POINTS TO STRESS 1. All assigned readings and exercises are from the textbook Objectives: Make certain that you can define, and use in context, the terms, concepts and formulas listed below: • Evaluate triple integrals in Cartesian Coordinates. In Rectangular Coordinates, the volume element, " dV " is a parallelopiped with sides: " dx ", " dy ", and " dz ". Cylindrical and Spherical Coordinates General substitution for triple integrals. For the calculation of areas, we use majorly integrals formulas. fdV≡ Triple integral of f over R dV = volume element in coordinate system which describes R. Definición de integral triple Una integral triple es una generalización de una integral doble en el mismo sentido que una doble es una generalización de una integral sencilla. Double Integrals Definition (1. » 2 2 » ? 4 y2 0 » ? 4 x2 y2? 4 x2 y2 x2 a x2 y2 z2 dzdxdy:. Triple Integrals in Cylindrical Coordinates Useful for circle-symmetrical integration regions and integrand functions Switch to polar coordinates for 2 of the 3 coordinates, leave the third as is x r cos y r sin z z f ( x, y , z ) f (r , , z ) dx dy dz r dr d dz Equivalent to integrate first inz , then in polar coordinates. ! Evaluate a double integral as an iterated integral. Preface This book covers calculus in two and three variables. Multivariable Calculus Seongjai Kim Department of Mathematics and Statistics Mississippi State University Mississippi State, MS 39762 USA Email: [email protected] Compute ½ T. It is merely another tool to help you get started studying. 1 Find the centroid of the solid that is bounded by the xz-plane and the hemispheresy= √ 9−x2 −z2 andy= √ 16−x2 −z2 assumingthedensityisconstant. the integral calculus courses. Cylindrical and Spherical Coordinates General substitution for triple integrals. It will come as no surprise that we can also do triple integrals—integrals over a three-dimensional region. Each ordering leads to a di erent description of the region of integration in space, and to di erent limits of integration. The first input, fun, is a function handle. c 2019 MathTutorDVD. Triple Integrals 3 Figuring out the boundaries of integration. Review of Chapter 16: Multiple Integrals Note: This review sheet is NOT meant to be a comprehensive overview of what you need to know for the exam. z = 8 − x 2 − y 2. •Triple Integrals can also be used to represent a volume, in the same way that a double integral can be used to represent an area. It has been given many names - the Project Management Triangle, Iron Triangle and Project Triangle - which should give you an idea of how important the Triple Constraint. Question: 4 otalT Credit 2 2 GPA Credit Points Earned. TRIPLE INTEGRALS 3 5B-2 Place the solid hemisphere D so that its central axis lies along the positive z-axis and its base is in the xy-plane. Ejemplo: Calcular la integral triple de f(x,y,z) = xy en la región definida por D = {( x , y , z ) Î R 3 | x 2 + y 2 + z 2 £ 1, x ³ 0, y ³ 0, z ³ 0. The solid below is enclosed by x= 0, x= 1, y= 0, z= 0, z= 1, and 2x+y+2z= 6. This means we'll write the triple integral as a double integral on the outside and a single integral on the inside of the form We'll let the -axis be the vertical axis so that the cone is the bottom and the half-sphere is the top of the ice cream cone. ) We will turn triple integrals into (triple) iterated integrals. In this video, I start discussing how a particular order of integration for a given region and integral ' makes sense '! Then I go. Changes of variable can be made using Jacobians in much the same way as for double integrals. Application is made to the case of an electrified disc with a hole in it and numerical results for the capacity of the. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. I (x − 2)2 + y2 = 4 is a circle, since. 321 Example 53. Notes on Triple integrals: Wednesday, November 26 These are some notes for my lecture on triple integrals. Find the φ-limits of integration. ) Write ZZZ U xyzdV as an iterated integral in cylindrical coordinates. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. We'll learn that integration and di erentiation are inverse operations of each other. I've uploaded my excel vba file. Integrals 6 1. Evaluating Triple Iterated Integrals. The following concepts may or may not be seen on the exam and there may be concepts on the exam which are not covered on this sheet. • M yz = RRR S xδ(x,y,z)dV is the moment about the yz-plane. Triple integration exercises 1. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. We used a double integral to integrate over a two-dimensional region and so it shouldn't be too surprising that we'll use a triple integral to integrate over a three dimensional region. Here we study double integrals Z Z Ω f(x;y)dxdy (5. 6 to find their infinite series forms. The Evaluation Theorem 11 1. To begin with, suppose that ˚(x;y;z) is a piecewise continuous function. INTEGRAL REVIEW July 2012 Vol. Others come from using di erent coordinate systems. z = 8 − x 2 − y 2. Write down all the conditions (boundary surfaces). 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. 7 Triple Integrals 4. Six of them can be obtained by permuting the order of the variables. 333 3 3 3 3 3 x dx x x x 4 32 1 5 5 5 5 75 4. We are given some solid region E in 3-space, and a function f(x,y,z), and we want to know 'how much of f is there in the region E'. Triple integral in cylindrical coordinates (Sect. Use cylindrical or spherical coordinates to evaluate the integral Z 1 1 Zp 1 2x 0 Zp 1 x2 y2 0 e (x 2+y2+z2)3=2dzdydx: Hint: Start by sketching the solid determined by the bounds of integration. Triple Integrals: A Hemisphere Example Let R be the region of three dimensional space bounded by z ≥0 and the surface of a sphere of radius a with a center at the origin. For any given θ, the angle φ that M makes with the z-axis runs from φ = φmin to φ = φmax. 6 Triple Integrals Comments. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. Fubini's theorem for triple integrals states that the value of a triple integral of a continuous function f over a region E in R 3 is a triple iterated integral. Lady (December 21, 1998) Consider the following set of formulas from high-school geometry and physics: Area = Width Length Area of a Rectangle Distance = Velocity Time Distance Traveled by a Moving Object Volume = Base Area Height Volume of a Cylinder Work = Force Displacement Work Done by a Constant Force. Triple integrals are the analog of double integrals for three dimensions. 14 Vector Equation of a Plane ~n(~r ~r 0) = 0 where ~nis the vector orthogonal to every vector in the given plane and ~r ~r. With these substitutions, the paraboloid becomes z=16-r^2 and the region D is given by 0<=r<=4 and 0<=theta<=2*pi. Problem 1: Set up the limits of integration for a triple integral ∫ ∫ ∫E f(x,y,z) dV where E ={(x,y,z)| 0 b x b 2, 1 b y b 2-x, 0 b z b 4- x 2 -y 2 }. 3 Triple Integrals Triple integrals of functions f (x , y, Z) of flu-ee variables are a fairly straightforward gen- erahzation of double integrals. The paper also summarizes the results of the survey questions given to the students in two of the courses followed by the authors own critique of the enhancement project. Change of Variables in Multiple Integrals: Euler to Cartan Author(s): Victor J. Convert each of the following to an equivalent triple integ4al. 14 Vector Equation of a Plane ~n(~r ~r 0) = 0 where ~nis the vector orthogonal to every vector in the given plane and ~r ~r. Integrals 6 1. So let us give here a brief introduction on integrals based on the Mathematics subject to find areas under simple curves, areas bounded by a curve and a line and area between two. The rectangular. Asymptotics of integrals of n-fold products We determine precise asymptotics in spectral parameters for integrals of n-fold products of zonal spherical harmonics on SL2(C). We'll learn that integration and di erentiation are inverse operations of each other. The simplest application allows us to compute volumes in an alternate way. Triple integration of sum of two functions is explained. 99 USD for 2 months 4 months:. Integrals with Trigonometric Functions Z sinaxdx= 1 a cosax (63) Z sin2 axdx= x 2 sin2ax 4a (64) Z sinn axdx= 1 a cosax 2F 1 1 2; 1 n 2; 3 2;cos2 ax (65) Z sin3 axdx= 3cosax 4a + cos3ax 12a (66) Z cosaxdx=. In the field of FEM, triple integrals need to be evaluated while finding the stiffness matrix, mass matrix, body force vector, etc. Convert each of the following to an equivalent triple integ4al. V = ∭ U ρ 2 sin θ d ρ d φ d θ. Engineering Mathematics III: UNIT I: Linear systems of equations: Rank-Echelon form-Normal form – Solution of linear systems – Gauss elimination – Gauss Jordon- Gauss Jacobi and Gauss Seidel methods. The double integral gives us the volume under the surface z = f(x,y), just as a single integral gives the area under a curve. (iv) Evaluate. Divide the cube into LxMxN small rectangular elements, each having. In particular, the minimum x-value occurs on the plane z= x+ 2 and the maximum xp-value occurs on the cylinder x 2+4y = 4. Integrals over V can be found as iterated integrals with the following result: FIGURE 1. donde Si es un punto en el interior de uno de estos bloques, entonces el volumen del bloque puede ser aproximado por. Don't show me this again. Math 232 Calculus III Brian Veitch Fall 2015 Northern Illinois University 15. The triple integral of f over the box B is lim E E AV if this limit exists. 4 (approximate answer, depends on what you estimated the values at the midpoints to be). Faraday's Law :. Integrals 6 1. Solution Figure 15. Integration by Parts 21 1. The deﬁnition and properties of the double integral. We could express the result in the equiv-alent form ZZZ D f(x,y,z)dxdydz = Z b 3 a3 ˆZZ R f(x,y,z)dxdy ˙ dz with f ≡ 1. Solution: First sketch the integration region. Notation: double integral of f over R= I f x y dxdy( , ) ³³ Note: Area element = dA = dx dy. (b) Let’s guess that this integral is divergent. Because if your integration order takes care of Z first, i. Setting up a Triple Integral in Spherical Coordinates. James McKernan, Maths, 18. Then came a second integral to add up the slices. 'tiled' integral3 calls integral to integrate over xmin ≤ x ≤ xmax. LECTURE 3: TRIPLE INTEGRALS (II) 5 In that case, the bigger function is the function in front, and the smaller one is the one in the back, and Dis the shadow behind the surface. The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals. Triple Integrals in Spherical Coordinates If you are studying an object with spherical symmetry, it makes sense to use coordinates to re ect that. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. The meaning of integration. TRIPLE INTEGRALS 385 4. LECTURE 3: TRIPLE INTEGRALS (II) 5 In that case, the bigger function is the function in front, and the smaller one is the one in the back, and Dis the shadow behind the surface. The usual “divide and conquer” approach for integrating f over B leads to the triple Riemann sum whose limit (if it exists) is the triple integral of f over B: ZZZ B f(x,y,z) dV = lim. c) Explain why any ordering starting with dz is not of Type I. Triple Integrals: A Hemisphere Example Let R be the region of three dimensional space bounded by z ≥0 and the surface of a sphere of radius a with a center at the origin. Triple Integrals over a General Bounded Region. Differential equations of first order and their. 7: Triple Integrals Outcome A: Evaluate a triple integral by iterated integration. patrickJMT 357,008 views. Math 232 Calculus III Brian Veitch Fall 2015 Northern Illinois University 15. Evaluating double integrals is similar to evaluating nested functions: You work from the inside out. the triple integral of f over the solid and denote it by RRR S f(x,y,z)dV. Try our award-winning software today. Partial Di erentiation and Multiple Integrals 6 lectures, 1MA Series Dr D W Murray Michaelmas 1994 Textbooks Most mathematics for engineering books cover the material in these lectures. First, lets describe the mass of a volume. For t2R, set F(t. Definición de integral triple Una integral triple es una generalización de una integral doble en el mismo sentido que una doble es una generalización de una integral sencilla. Cylindrical and Spherical Coordinates General substitution for triple integrals. Compute volume, mass, and center of mass using triple integrals. We partition a rectangular boxlike region containing D into. Triple Integrals in Cylindrical/Spherical Coordinates Multi-Variable Calculus. We shall use the following standard definitions for Laguerre polynomials (1) and Laguerre functions (2): (2) X„(x) = e"l/2L„(x) The Laguerre functions are known to constitute a complete orthonormal set in L2(0, a> ). pdf from MATH 267 at University of Calgary. Let us divide D into n subregions di whose areas are equal to si, choose a point (ξ i. 34 videos Play all MULTIPLE INTEGRALS (Complete Playlist) MKS TUTORIALS by Manoj Sir Triple Integrals, Changing the Order of Integration, Part 1 of 3 - Duration: 12:52. The limits for z arise from expressing the equation for the surface of the ellipsoid in the form z= c. 5: Triple Integrals. Triple integrals also arise in computation of Volume (if f(x,y,z)=1, then the triple integral equals the volume of R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia Triple Integrals in General Regions. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. 1 The Double Integral over a Rectangle Let f = f(x, y) be continuous on the Rectangle R: a < x < b, c < y < d. The double integral of a nonnegative function f(x;y) deﬂned on a region in the plane is associated with the volume of the region under the graph of f(x;y). Just as for double integrals, a region over which a triple integral is being taken may have easier representation in another coordinate system, say in uvw-space, than in xyz-space. Triple integral of "height" w = f(x,y,z) times infinitesimal volume = total 4d hypervolume under 3d region. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). Warning: This is only possible if all the limits of integration are numbers and the integrand is completely separable as a prod-uct of functions of a single variable. The corresponding multiple integrals are referred to as double integrals, triple integrals, and n-tuple integrals, respectively. 1 Structure and Process: Integral Philosophy and Triple Transformation Debashish Banerji1 Abstract: This paper looks at the ongoing debate between perennialism and pluralism in religious studies and considers the category of the integral, as described by Sri Aurobindo. The infinite series forms of the two types of triple integrals can be obtained using binomial series and integration term by term theorem. 7) Example Use cylindrical coordinates to ﬁnd the volume in the z > 0 region of a curved wedge cut out from a cylinder (x − 2)2 + y2 = 4 by the planes z = 0 and z = −y. The simplest application allows us to compute volumes in an alternate way. Each ordering leads to a di erent description of the region of integration in space, and to di erent limits of integration. Now we define triple integrals for functions of three variables. • If δ(x,y,z) is the density of the solid at the point (x,y,z), then M = RRR S δ(x,y,z)dV gives the mass of the solid. 1, Introduction to Determinants In this section, we show how the determinant of a matrix is used to perform a change of variables in a double or triple integral. Multiple integrals possess a number of properties similar to those. boundary surface of E is equal to the triple integral of the divergence of F over E. (Note: The paraboloids intersect where z= 4. donde Si es un punto en el interior de uno de estos bloques, entonces el volumen del bloque puede ser aproximado por. Triple Integrals in Rectangular Form Note. pdf - Google Drive Sign in. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates. We illustrate with some examples. TRIPLE INTEGRALS EXTRA CREDIT FOR MATH 222-01/02 DUE NOVEMBER 16, 2011 The answers to the problems below should be presented neatly, (either typed or written very neatly). We can compute R fdA on a region R in the following way. Because if your integration order takes care of Z first, i. The first variable given corresponds to the outermost integral and is done last. with respect to each spatial variable). Math symbols deﬁned by LaTeX package «esint» No. 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. Aplicaciones de de la integral Volumen de sólidos de revolución Definición Sea una función definida en el intervalo. Notice how the inequalities involve xand y. Triple integral is defined and explained through solved examples. above z = −√4x2 +4y2. Triple integral of infinitesimal volume = total volume of 3d region. I Project your region E onto one of the xy-, yz-, or xz-planes, and. Cylindrical and Spherical Coordinates General substitution for triple integrals. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). The copyright holder makes no representation about the accuracy, correctness, or. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Let x i = x i x i1 be the width of the i’th subinterval [x i1,x i] and let the norm of the partitionkPkbethelargestofthex i’s. For a triple integral in a region D ˆR3, it can be evaluated by using an iterated. The simplest application allows us to compute volumes in an alternate way. Look for a variable that has. Sometimes we can reduce a very diﬃcult double integral to a simple one via a substitution. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. If f is continuous, the triple integral exists and does not depend on the choice of (x∗ ijk,y ∗ ijk,y ∗ ijk) Same properties as double integrals Mth 254H – Fall 2006 3/10 Evaluating Fubini’s Theorem: If f(x,y,z) is continuous on B =[a,b]×[c,d]×[r,s], then ZZZ R f(x,y,z)dV = Z s r Z. This description is too narrow: it's like saying multiplication exists to find the area of rectangles. It can also evaluate integrals that involve exponential, logarithmic, trigonometric, and inverse trigonometric functions, so long as the. Sketch the solid whose volume is given by the iterated integral. 3 Find the volume of the region bounded by z = 50 10y, z = 10, y = 0, and y = 4 x2. Triple integral of infinitesimal volume = total volume of 3d region. 1 2x 2y 2= 1 2(x + y) = 1 r 2. Fubini's theorem for triple integrals states that the value of a triple integral of a continuous function f over a region E in R 3 is a triple iterated integral. Definición de integral triple Una integral triple es una generalización de una integral doble en el mismo sentido que una doble es una generalización de una integral sencilla. Express the big integral like that, and evaluate each single integral separately. It will be mostly about adding an incremental process to arrive at a \total". above z = −√4x2 +4y2. The assignment is due at the beginning of class, August 6th. We'll learn that integration and di erentiation are inverse operations of each other. The triple integral of a continuous function over a general solid region. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). 4 EXERCISES Review Questions 1. Read Section 16. I've uploaded my excel vba file. History of the Integral from the 17 th Century. 34 videos Play all MULTIPLE INTEGRALS (Complete Playlist) MKS TUTORIALS by Manoj Sir Triple Integrals, Changing the Order of Integration, Part 1 of 3 - Duration: 12:52. Integrals of a function of two variables over a region in R 2 are called double integrals, and integrals of a function of three variables over a region of R 3 are called triple integrals. Challenge: 11,23 4. S8: Double integrals in polar co–ordinates. Double integrals in polar coordinates. Just as the definite integral of a positive function of one variable represents the area of the region between the. MULTIPLE INTEGRALS 154 15. \mathbf {F} = – Gm\,\mathbf {\text {grad}}\,u, where G is the gravitational constant. stackexchange [22], and in a slightly less elegant form it appeared much earlier in [18]. x y z x + y = 12 2 z = 1 - x - y 2 2 141. Engineering Mathematics 1 Notes Pdf – EM 1 Notes Pdf UNIT – V. Read Section 16. Triple Integrals in Spherical Coordinates If you are studying an object with spherical symmetry, it makes sense to use coordinates to re ect that. 1 Multiple-Integral Notation Previously ordinary integrals of the form Z J f(x)dx = Z b a f(x)dx (5. When we have to integrate a function of x,y,z over all space, we write a triple integral in this way: ∫ − ∞ + ∞ ∫ − ∞ + ∞ ∫ − ∞ + ∞ (,,). Approximating Integrals In each of these cases, the area approximation got better as the width of the intervals decreased. 2 Assignments 1. It's difficult to explain. Our first integral could equally well be ff(x, y)dx. Contents 1. Then using expan-sions of K and K he had obtained himself 30 years previously—Watson [6]—he was able to evaluate these last integrals. Multiple Integrals Double Integrals over Rectangles 26 min 3 Examples Double Integrals over Rectangles as it relates to Riemann Sums from Calc 1 Overview of how to approximate the volume Analytically and Geometrically using Riemann Sums Example of approximating volume over a square region using lower left sample points Example of approximating volume over a…. Sam Johnson (NIT Karnataka) Triple Integrals in Rectangular Coordinates October 24, 2019 17/62. (a) » 2 0 » 1 0 » 3 0 p xy z2q dzdydx (b) » 2 0 » z2 0 » y z 0 p 2x yq dxdydz (c) » ˇ{ 2 0 » y 0 » x 0 cosp x y zq dzdxdy 2. However each student is responsible. Fill in the limits of integration for the integral and put the dx, dy, dz in the correct order:. Therefore, the desired function is f(x)=1 4. Triple Integrals, Changing the Order of Integration, Part 1 of 3. All these methods are Numerical. The cylindrical coordinate system describes a point (x,y,z) in rectangular space in terms of the triple (r,θ,z) where r and θ are the polar coordinates of the projection. Welcome! This is one of over 2,200 courses on OCW. Both of the limits diverge, so the integral diverges. The analogy between single and. We used a double integral to integrate over a two-dimensional region and so it shouldn't be too surprising that we'll use a triple integral to integrate over a three dimensional region. Although the prerequisite for this. Triple Integrals: A Hemisphere Example Let R be the region of three dimensional space bounded by z ≥0 and the surface of a sphere of radius a with a center at the origin. In addition, some examples are used to demonstrate the calculations. INTEGRAL REVIEW July 2012 Vol. The corresponding multiple integrals are referred to as double integrals, triple integrals, and n-tuple integrals, respectively. and above the region in the xy. INTEGRALS 289 Thus, {F + C, C ∈ R} denotes a family of anti derivatives of f. Look for a variable that has. iosrjournals. Let Ube the solid inside both the cone z= p. Integrals are often described as finding the area under a curve. 3 0 2ˇ 0 2 1 (r+ z)rdrd dz Region from Diagram 2 3 0 2ˇ 0 2 0 5zrdrd dz Region from Diagram 2 3 0 ˇ=2 0 2 1. Convert integrals into another form For a double integral in a region D ˆR2, it can be evaluated by using an iterated integral in dxdy, or dydx, or using polar coordinate. If you want to calculate the area under the curve or some definite integral in the Symbolic (Analytical) way, then it is very hard to using C++ and not very useful. We'll learn that integration and di erentiation are inverse operations of each other. Z 1 1 1 + e x x dx Solution: (a) Improper because it is an in nite integral (called a Type I). Solution: ZZ D (x +y)dA = Z1 0 Z2y −y (x+y)dxdy = Z1 0 (x2 2 +xy) x=2y x=−y = Z1 0 9y2 2 dy = 3y3 2 y=1 y=0 = 3 2. The multiple integral is a type of definite integral extended to functions of more than one real variable—for example, $f(x, y)$ or $f(x, y, z)$. Imagine you have a cube that's gets denser as you move further out towards its corners. Triple Integrals over a General Bounded Region. Integral calculus that we are beginning to learn now is called integral calculus. Una integral triple es una generalización de una integral doble en el mismo sentido que una doble es una generalización de una integral sencilla. Elementary Approach to the Dirichlet Integral 1 2. Dirichlet integral, is often evaluated using complex-analytic methods, e. Integration Method Description 'auto' For most cases, integral3 uses the 'tiled' method. (We just add a third dimension. Free triple integrals calculator - solve triple integrals step-by-step. Evaluate the triple integral. This article is about the Euler–Poisson integral. Integrales Triples Hermes Pantoja Carhuavilca3 de 30. Double Integrals Definition (1. The line integral of a magnetic field around a closed path C is equal to the total current flowing through the area bounded by the contour C (Figure 2). patrickJMT 357,008 views. Homework 6 Solutions Jarrod Pickens 1. This sum has a nice interpretation. Engineering Mathematics III: UNIT I: Linear systems of equations: Rank-Echelon form-Normal form - Solution of linear systems - Gauss elimination - Gauss Jordon- Gauss Jacobi and Gauss Seidel methods. 8 The Fubini’s Method to evaluate triple integrals is this: We ﬁrst walk on a line parallel to one of the x-axis, y-axis, or z-axis. Triple Integral Practice To Set Up A Triple Integral 1. Example Use cylindrical coordinates to ﬁnd the volume of a curved wedge cut out from a cylinder (x − 2)2 + y2 = 4 by the planes z = 0 and. 7 We integrate the triple integral directly. Triple Integrals, Changing the Order of Integration, Part 1 of 3. and inside x2 +y2 = 4. Single Integral - the domain is the integral I (a line). set up the triple integral in terms of single integrals, but do not evaluate it). Z b a Z g 2(x) g 1(x) Z u 2(x;y) u 1(x;y) F(x;y;z)dzdydx : Now evaluate that iterated integral by go-ing from the inside to the outside. Example Use cylindrical coordinates to ﬁnd the volume of a curved wedge cut out from a cylinder (x − 2)2 + y2 = 4 by the planes z = 0 and. Usually, one direction is considered to be positive, the other negative. To approximate a volume in three dimensions, we can divide the three-dimensional region into small rectangular boxes, each \(\Delta x\times\Delta y\times\Delta z. Problems: 5,7,9,13,17,33 3. Step 1: Draw a picture of E and project E onto a coordinate plane. Triple integral is an integral that only integrals a function which is bounded by 3D region with respect to infinitesimal volume. Triple Integrals in Cylindrical Coordinates A point in space can be located by using polar coordinates r,θ in the xy-plane and z in the vertical direction. Use the Comparison Theorem to decide if the following integrals are convergent or divergent. Triple integrals Triple integral examples 3c. Try our award-winning software today. To begin with, suppose that ˚(x;y;z) is a piecewise continuous function. Partial Di erentiation and Multiple Integrals 6 lectures, 1MA Series Dr D W Murray Michaelmas 1994 Textbooks Most mathematics for engineering books cover the material in these lectures. The two integrals that have dy as the innermost di erential are Z2 0 Zx 0 x 0 ex(y + 2z) dydzdx + 2 0 Z2x x x z x ex(y + 2z. The simplest application allows us to compute volumes in an alternate way. A solid region Eis said to be of type 1 if it lies between. Evaluating triple integrals A triple integral is an integral of the form Z b a Z q(x) p(x) Z s(x,y) r(x,y) f(x,y,z) dzdydx The evaluation can be split into an "inner integral" (the integral with respect to z between limits. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. (a)Set up but do not evaluate a single triple integral to nd the volume of Susing cylindrical coordinates. Math 21a Triple Integrals Fall, 2010 1 Evaluate the integral RRR E 2xdV, where E= {(x,y,z) : 0 ≤y≤2,0 ≤x≤ p 4 −y2,0 ≤z≤y}. b) Set up a triple integral over S in the dx dy dz ordering. The triple integral of a continuous function over a general solid region. We'll learn that integration and di erentiation are inverse operations of each other. Abraham de Moivre originally discovered this type of integral in 1733, while Gauss published the precise integral in 1809. This depends on finding a vector field whose divergence is equal to the given function. Accordingly, its volume is the product of its three sides, namely dV dx dy= ⋅ ⋅dz. Triple Integral Spherical Coordinates Pdf Download >>> DOWNLOAD (Mirror #1). Numerical Integration and Differentiation Quadratures, double and triple integrals, and multidimensional derivatives Numerical integration functions can approximate the value of an integral whether or not the functional expression is known:. 1 DOUBLE INTEGRALS OVER RECTANGLES TRANSPARENCIES AVAILABLE #48 (Figures 4 and 5), #49 (Figures 7 and 8), #50 (Figure 11), #51 (Figures 12 and 13) SUGGESTED TIME AND EMPHASIS 1 2 –1 class Essential Material POINTS TO STRESS 1. V = \iiint\limits_U {\rho d\rho d\varphi dz}. The cylindrical coordinate system describes a point (x,y,z) in rectangular space in terms of the triple (r,θ,z) where r and θ are the polar coordinates of the projection. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 5 3. Challenge: 11,23 4. Applications of Double/Triple Integrals. Such integrals arise whenever two functions are multiplied, with both the operands and the result represented in the Legendre polynomial basis. dimensional integrals. dimensional domain. Just as for double integrals, a region over which a triple integral is being taken may have easier representation in another coordinate system, say in uvw-space, than in xyz-space. Try to visualize the 3D shape if you can. P 1 f(P1)=14kg=m3 f(P2)=7kg=m3 P 2 P 3 f(P3)=9kg=m3 P 4 f(P4)=21kg=m3 Suppose f 2C(R3) measures density (kg=m3) throughout W. To obtain double/triple/multiple integrals and cyclic integrals you must use amsmath and esint (for cyclic integrals) packages. 5 36 Triple Integral Strategies The hard part is guring out the bounds of your integrals. Let D be the half-washer 1 x2 + y2 9, y 0, and let E be the solid region above D and below the graph z = 10 x2 y2. Integrals 6 1. where d is the radius of rotation. The divergence theorem can also be used to evaluate triple integrals by turning them into surface integrals. (5), if ,, then the triple. James McKernan, Maths, 18. Triple Integrals | x12. I Project your region E onto one of the xy-, yz-, or xz-planes, and use the boundary of this projection to nd bounds on domain D. We can interpret this result as the volume of the solid region because the integrand is 1. Aplicaciones de de la integral Volumen de sólidos de revolución Definición Sea una función definida en el intervalo. Triple integration of sum of two functions is explained. -plane defined by 0 ≤ x ≤ 2. This article considers two types of triple integrals and uses Maple for verification. z = 8 − x 2 − y 2. an integral of a function defined on some region in a plane and in three-dimensional or n-dimensional space. Notice how the inequalities involve xand y. Solution: We'll use the shadow method to set up the bounds on the integral. By using this website, you agree to our Cookie Policy. stackexchange [22], and in a slightly less elegant form it appeared much earlier in [18]. (1) is deﬂned as Z C a ¢ dr = lim N!1 XN p=1 a(xp;yp;zp) ¢ rpwhere it is assumed that all j¢rpj ! 0. Use a triple integral to determine the volume of the region below z = 4−xy. So then x2 +y2 = r2. Multiple integrals use a variant of the standard iterator notation. Challenge Problems. 25 3 4 3 12 4 tt t t dt 1. via contour integration. (Or vice versa. EXAMPLE 4 Find a vector field whose divergence is the given F function. MTH 254 LESSON 20. Triple Integral Calculator Added Mar 27, 2011 by scottynumbers in Mathematics Computes value of a triple integral and allows for changes in order of integration. The triple integral in this case is, Note that we integrated with respect to x first, then y, and finally z here, but in fact there is no reason to the integrals in this order. So let us give here a brief introduction on integrals based on the Mathematics subject to find areas under simple curves, areas bounded by a curve and a line and area between two. In Eastern Europe, it is known as Ostrogradsky’s. Notation: double integral of f over R= I f x y dxdy( , ) ³³ Note: Area element = dA = dx dy. 2, we showed how a function of two variables can be inte-grated over a region in 2-space and how integration over a region is equivalent to an iterated or double integral over two intervals. If we substitute back into the sum we get nX−1 i=0 G(yi)∆y. Recall that if you're looking for th volume then f. Hence h 1(y;z) = z 2 and h 2(y;z) = 4 4y2 = 2 p 1 y2. We would like to be able to integrate triple integrals for more general regions. pdf - Free download as PDF File (. 1 De nition of double integral Consider the function of two variables f(x,y) deﬁned in the bounded region D. Hence the triple integral becomes Z Z Z D dV = Z 4 0 Z 1 1 Z 2 p 1 y2 z 2 1 dxdydz = Z 4. 6 to find their infinite series forms. fun(x,y,z) must accept a vector x and scalars y and z, and return a vector of values of the integrand. A solid region Eis said to be of type 1 if it lies between. Find the θ-limits of integration. However each student is responsible. We'll learn that integration and di erentiation are inverse operations of each other. Note, that integral expression may seems a little different in inline and display math mode - in inline mode the integral symbol and the limits are compressed. 5-8: Surface Area, Triple Integrals Friday, April 8 Surface Area Using the formula A(S) = ZZ D q 1 + f2 x + f y 2 dA, nd the surface area of a sphere of radius a. In this video, I start discussing how a particular order of integration for a given region and integral ' makes sense '! Then I go. V = ∭ U ρ 2 sin θ d ρ d φ d θ. Don't show me this again. By symmetry, ¯x = 0 and ¯y = 0, so we only need ¯z. Integrales Triples Hermes Pantoja Carhuavilca3 de 30. Triple integrals also arise in computation of Volume (if f(x,y,z)=1, then the triple integral equals the volume of R) Force on a 3D object Average of a Function over a 3D region Center of Mass and Moment of Inertia Triple Integrals in General Regions. Triple integrals are the analog of double integrals for three dimensions. Triple Integrals in Cylindrical/Spherical Coordinates Multi-Variable Calculus. Find materials for this course in the pages linked along the left. This website uses cookies to ensure you get the best experience. The meaning of integration. use the trapezoidal rule of integration to solve problems, 3. (iv) Evaluate. » Integrate can evaluate integrals of rational functions. In physics, triple integral arises in the computation of mass, volume, moment of inertia and force on a three dimensional object. If we want to convert this triple integral to cylindrical coordinates we need to rewrite xand yusing the conversion formulas from above. Triple Integrals Motivation Example An object conforms to the shape of a solid W in R3. Imagine you have a cube that's gets denser as you move further out towards its corners. (b) Let's guess that this integral is divergent. The solid below is enclosed by x= 0, x= 1, y= 0, z= 0, z= 1, and 2x+y+2z= 6. (Q9)Set up the triple integral W dV = W 1dV , using the order of integration dV = dz dy dx. 7 Triple Integrals in Cylindrical and Spherical Coordinates Example: Find the second moment of inertia of a circular cylinder of radius a about its axis of symmetry. 6 to find their infinite series forms. Compute the following integral by making a change in coordinates. Step 2: Determine the limits of integration. It is also useful in setting up triple integrals as iterated integrals to let Rbe the. Accordingly, its volume is the product of its three sides, namely dV dx dy= ⋅ ⋅dz. patrickJMT 357,008 views. The sphere x2 +y2 +z2 = 4 is the same as ˆ= 2. (Unfortunately, it's harder to draw in three dimensions. Solution: a) Sketch an arrow in the positive y direction: This arrow enters the solid at the xz-plane ( 1=0), passes through the interior (gray), and. 4 EXERCISES Review Questions 1. The Evaluation Theorem 11 1. This week’s review talks about Triple Integrals and Applications. Double Integrals Definition (1.
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## anonymous one year ago The question 1B-6?

1. phi

can you post the question?

2. phi

is it what is the locus of points for the equation $\vec{OP} \cdot \textbf{u}= c | \vec{OP}|$ ?

3. anonymous

yes, I can't understand: "Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ" How can I reach this conclusion

4. anonymous

1B-6 Let O be the origin, c a given number, and u a given direction (i.e., a unit vector). Describe geometrically the locus of all points P in space that satisfy the vector equation OP · u = c | OP | . In particular, tell for what value(s) of c the locus will be (Hint: divide through by | OP | ): a) a plane b) a ray (i.e., a half-line) c) empty

5. phi

if we divide both sides by |OP| we get $\frac{OP}{|OP|} \cdot u = c$ that is two unit vectors. the maximum value of v dot u is 1 when v and u are unit vectors

6. phi

now consider c=0 any vector OP perpendicular to u , when dotted with u will give 0 |dw:1439222905180:dw|

7. phi

any point P in the plane defined by u (as the normal) will form vector OP, and after dividing by |OP| will still be a vector (of unit length) in that plane. and dotted with u, will give 0 thus c=0 results in a plane

8. phi

if |c| = 1 (c=1 or c=-1) the locus will be a ray if c=1 then any point P where O to P is in the same direction as u will, after being divided by the length of OP will give us unit vector v = u and u dot v = 1

9. anonymous

If we applied the equation of Dot Product: $\left| A \right| \left| B \right| \cos \theta = A.B = \sum_{i }^{...} a _{i}.b_{i}$ W ell, u is a unit vecton, then $\left| u \right| = 1$ then, $\frac{ OP.u }{ \left| OP \right|}=\cos \theta = c$ Off course, this is the angle between these two vector, The points P need stay

10. anonymous

points P are at $\theta$ of in relation to vectior u. But and when the corretion says: Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ 0 . I can understand.

11. phi

for 0 < |c| < 1 we get a cone

12. anonymous

Ok Thanks Phi, I will see it carefully, thanks by help me!

13. phi

what part of the answer is not clear?

14. phi

***"Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ" How can I reach this conclusion**** |dw:1439223719035:dw|

15. phi

let c= 0.5 and OP/|OP| = v v dot u = 0.5 theta = 60 degrees any point P so that OP is in the direction 60 degrees away from u (in any direction) will be scaled to unit length. and then dotted with u = 0.5

16. anonymous

Ok, Itś clear now! Thaks... Now I will remake this exercise.

17. anonymous

Unfortunately I can´t get the concept on the last part of this question, c) empty response: Locus is the origin, if c > 1 or c < −1 (division by | OP | is illegal, notice).

18. anonymous

it is just because cos or sen modules can't be larger than 1. It is | sen(x) | <=1 and |cos(x)|<=1 ?

19. phi

yes, I was confused on that for a bit. if we want $\frac{OP}{|OP|} \cdot u = 2$ for all P not at the origin, the vector OP divided by its length will give us a unit vector, call it v; and v dot u =2 will never be true. I originally thought this meant "no solution" but if P is the origin, then OP is the zero length vector and $\vec{OP} \cdot \textbf{u} = 2| OP|$ becomes $\vec{OP} \cdot \textbf{u} = 2\cdot 0= 0$ or $<0,0,0> \cdot <a,b,c> = 0$ where I am assuming we are in 3D space.

20. phi

and <0,0,0> dot u gives 0 therefore the zero vector $$\vec{\textbf{0}}$$ is a solution for |c|>1

21. phi

v dot u =2 to elaborate v dot u = |u| |v| cos A because |u| and |v| are both 1 (both are unit vectors) and the max value of cos A is 1 the max value for v dot u is 1 we can never get a value larger than 1

22. anonymous

Ok, Let's a simplest way to show "a vector divided by its length" is a unit vector.... I used matlab... >> OP = [-4 -45 1] OP = -4 -45 1 >>mag=sqrt(4^2+45^2+1^2) mag = 45.1885 >>OP_unitVector=OP./mag OP_unitVector = -0.0885 -0.9958 0.0221 mag_unitVector=sqrt(0.0885^2 + 0.9958^2 + 0.0221^2 ) mag_unitVector = 1.0000

23. phi

we can do it algebraically say we have v= <a,b,c> then $$|v|= \sqrt{a^2+b^2 + c^2}$$ (which follows from pythagoras) and $|v|^2 = a^2 + b^2 + c^2$ $\frac{v}{|v|} \cdot \frac{v}{|v|}= \frac{v \cdot v}{|v|^2}$ $v \cdot v= a^2 + b^2 + c^2$ and therefore $\frac{v \cdot v}{|v|^2} = \frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1$

24. phi

if we let $u = \frac{v}{|v|}$ we have shown $u \cdot u = |u|^2= 1 \\ \sqrt{|u|^2} = \sqrt{1}\\ |u|= 1$ and therefore $\Bigg|\frac{v}{|v|}\Bigg|= 1$ ie. is unit length


# Prove the following inequality: $\int_{0}^{\frac{\pi }{2}}\frac{sin(x)}{\sqrt{9-sin^{4}(x)}}dx\geq \frac{1}{3}$

Prove the following inequality: $$\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sqrt{9-\sin^{4}x}}\ \mathrm dx\geq\frac{1}{3}.$$

I am thinking of replacing the equation with $$\int_{0}^{\frac{\pi }{2}}\frac{\sin x}{9-\sin^{4} x }dx\geq \frac{1}{3}$$, however I am stuck at this point.

Do you have any suggestions?

• use the fact $\sqrt{9 - \sin^4x} \leq 3$ and $\int_0^{\pi/2} sin x dx = 1$ and $\sin x$ is nonegative on $[0,\pi/2]$. Feb 5 '20 at 18:07
• You can get the proper font and spacing for $\sin$ using \sin. For operators that don't have a command of their own, you can use \operatorname{name}. Feb 5 '20 at 18:19

\begin{align*}\int_{0}^{\pi/2} \frac{\sin{x}}{\sqrt{9-\sin^4{x}}}dx &\geq \int_{0}^{\pi/2}\frac{\sin{x}}{\sqrt{9}}dx\\& = \frac{1}{3}\int_{0}^{\pi/2} \sin{x}dx \\&=\frac{1}{3}[-\cos{\pi/2}-(-\cos{0})]\\&=\frac{1}{3}[0+1]\\&=\frac{1}{3} \end{align*}
The inequality is because $$\sqrt{9-\sin^4{x}}\leq \sqrt{9}$$ due to positivity of $$\sin^4{x}$$. So ... $$\frac{1}{\sqrt{9-\sin^4{x}}}\geq \frac{1}{\sqrt{9}}$$
• It's also $< 1/(2\sqrt{2})$. Feb 5 '20 at 20:23
The given inequality can be deduced from AM-GM. Indeed$$\begin{eqnarray*} I=\int_{0}^{\pi/2}\frac{\sin x}{\sqrt{9-\sin^4 x}}\,dx &=& \int_{0}^{1}\frac{z\,dz}{\sqrt{(9-z^4)(1-z^2)}}\\ &=& \frac{1}{2}\int_{0}^{1}\frac{du}{\sqrt{(3-u)(3+u)(1-u)}}=\int_{0}^{1}\frac{dv}{\sqrt{(2+v^2)(4-v^2)}}\end{eqnarray*}$$ and over $$[0,1]$$ we have $$\sqrt{(2+v^2)(4-v^2)}=\text{GM}(2-v^2,4+v^2)\leq\text{AM}(2-v^2,4+v^2)=3$$.
This also shows that $$I$$ is an incomplete elliptic integral of the first kind.


Idempotents in $\mathbb Z_n$

An element $$a$$ of the ring $$(P,+,\cdot)$$ is called idempotent if $$a^2=a$$. An idempotent $$a$$ is called nontrivial if $$a \neq 0$$ and $$a \neq 1$$.

My question concerns idempotents in rings $$\mathbb Z_n$$, with addition and multiplication modulo $$n$$, where $$n$$ is natural number. Obviously when $$n$$ is a prime number then there is no nontrivial idempotent. If $$n$$ is nonprime it may happen, for example $$n=4, n=9$$, that also there is no.

Is it known, in general, for what $$n$$ there are nontrivial idempotents and what is a form of such idempotents?

• They certainly do occur: $n=6,a=3.$ – Andrew Jul 27 '12 at 19:07
• There is a nontrivial idempotent in $\mathbb{Z}_n$ if and only if there is $1<a<n$ such that $n\, |\, a(a-1)$. This is because $a^2 \equiv a \pmod n \Leftrightarrow a^2-a \equiv 0 \pmod n$. I'm not entirely sure where you can go with this though! – Clive Newstead Jul 27 '12 at 19:14
• @CliveNewstead Since $a$ and $a-1$ are relatively prime, you get that such $a$ exists if and only if $n$ has two relatively prime nontrivial divisors, and this is equivalent to $n$ having two distinct primes... – N. S. Jul 27 '12 at 19:31

5 Answers

As often happens when dealing with $$\mathbf{Z}_n$$, the Chinese remainder theorem is your friend. If the prime factorization of $$n$$ is $$n=\prod_i p_i^{a_i},$$ then by CRT we have an isomorphism of rings $$\mathbf{Z}_n\cong\bigoplus_i \mathbf{Z}_{p_i^{a_i}}.$$ Observe that the isomorphism maps the residue class of an integer $$m$$ (modulo $$n$$) to a vector with all the components equal to the residue class of $$m$$ (this time modulo various prime powers): $$\overline{m}\mapsto(\overline{m},\overline{m},\ldots,\overline{m}).$$ So the residue class of $$m$$ is an idempotent if and only if it is an idempotent modulo all the prime powers $$p_i^{a_i}$$.

Let us look at the case of a prime power modulus $$p^t$$. The congruence $$x^2\equiv x\pmod{p^t}$$ holds, iff $$p^t$$ divides $$x^2-x=x(x-1)$$. Here only one of the factors of, $$x$$ or $$(x-1)$$, can be divisible by $$p$$, so for the product to be divisible by $$p^t$$ the said factor then has to be divisible by $$p^t$$. Thus we can conclude that $$x\equiv 0,1 \pmod{p^t}$$ are the only idempotents modulo $$p^t$$. Therefore we require that $$m\equiv 0,1\pmod{p_i^{a_i}}$$ for all $$i$$. By CRT these congruences are independent for different $$i$$, so the number of pairwise non-congruent idempotents is equal to $$2^\ell$$, where $$\ell$$ is the number of distinct prime factors $$p_i$$ of $$n$$.

• Can you please explain why $m\equiv 0,1\pmod{p_i^{a_i}}$ ? I understood everything up till that point – Belgi Jul 28 '12 at 0:55
• @Belgi, the notation was somewhat inconsistent, and the logic a bit of patchwork. Is it clearer now? – Jyrki Lahtonen Jul 28 '12 at 4:48
• What is "m"? ... – Stephen Mar 12 '15 at 19:32
• @Stephen: An integer. More precisely, an integer such that its residue class is an idempotent as explained on the sixth line. – Jyrki Lahtonen Mar 12 '15 at 19:35
• Sorry about the bump. I was reviewing some of my old answers and felt that this could be made a lot clearer (particular in view of Stephen's comment). Hopefully I succeeded. – Jyrki Lahtonen Aug 24 '18 at 9:51

Hint  Idempotents in $$\:\Bbb Z_{ n}\:$$ correspond to factorizations of $$\:n\:$$ into two coprime factors. Namely, if $$\:e^2 = e\in\Bbb Z_n\:$$ then $$\:n\:|\:e(e\!-\!1)\:$$ thus $$\:n = jk,\ j\:|\:e,\ k\:|\:e\!-\!1,\:$$ so $$\:(j,k)= 1\:$$ by $$\:(e,e\!-\!1) = 1.\:$$ Conversely if $$\:n = jk\:$$ for $$\:(j,k)= 1,\:$$ then by CRT, $$\:\Bbb Z_n\cong \Bbb Z_j\times \Bbb Z_k\:$$ which has nontrivial idempotents $$\:(0,1),\,(1,0).\:$$ It is not that difficult to explicitly work out the details of the correspondence. Some integer factorization algorithms search for such nontrivial idempotents.

This correspondence between idempotents and factorizations holds more generally at the structural level - see the Peirce Decomposition.

• It's worth noting here that if $x$ is an idempotent corresponding to a given $j$ and $k$, then $(p+)1-x$ is the other one, so it suffices to compute the idempotent corresponding to either $(0,1)$ or $(1,0)$. – rogerl May 29 '15 at 15:17
• @rogerl Yes that is clear since $\,(0,1)+(1,0) = (1,1) = 1\ \$ – Bill Dubuque Jun 14 '15 at 20:54

By the Chinese remainder theorem $\mathbf Z/n\mathbf Z$ is a product of more than one (unital) ring if and only if $n$ has more than one prime factor, and in this case $\mathbf Z/n\mathbf Z$ certainly has nontrivial idempotents. If on the other hand $n=p^k$ is a prime power, then all elements are either invertible (if not divisible by $p$) or nilpotent (if divisible by $p$) and this excludes the possibility of nontrivial idempotents. The case $n=1$ is a special case of a prime power (but the unique element now is both invertible and nilpotent; there are still no nontrivial idempotents of course).

A start: You can figure it out! Let us start with a product $mn$ of relatively prime integers, neither equal to $1$. By the Chinese Remainder Theorem, there is an $x$ such that $x\equiv 0\pmod{m}$ and $x\equiv 1 \pmod{n}$. Then $x^2\equiv x \pmod{mn}$.

Generalize to a product of $k$ relatively prime integers. If you use the prime power factorization, you can get a complete list.

Let $m=p^{c_{1}}_{1}...p^{c_{n}}_{n}$ be a prime factorization of an integer $m$ with $c_{i}\geq1$ and $p_{i}$ are distinct prime numbers. Then the ring $\mathbb{Z}/m\mathbb{Z}$ has $2^{n}$ idempotents and each one is precisely of the form $\sum\limits_{k=1}^{n}h_{k}\epsilon_{k}+m\mathbb{Z}$ where $\epsilon_{k}\in\{0,1\}$ and $h_{k}\in(\prod\limits_{\substack{i=1,\\ i\neq k}}^{n}p^{c_{i}}_{i})\mathbb{Z}$ such that $h_{k}-1\in p^{c_{k}}_{k}\mathbb{Z}$.


# I Affine Subspace

1. Jun 14, 2017

### Erik109

Hey,

I am struggling with developing an intuition behind 'Affine Subspaces'. So far I have read the theories concerning Affine Subspaces delivered by the course book and visited several websites, however none have made it 100% clear. I feel like I have some sort of intuition for it, but I fail to apply the intuition when it comes to solving problems.

At the moment, I am often required to show why a given set is an Affine Subspace of a certain space. Because I assume it is quite hard for you to convey the intuition behind it without writing a lot, I will try to convey the way I think by approaching a sample problem. I hope you can enlighten me on the errors/misconceptions I make or perhaps add something so I can learn more.

The sample problem:

Let X1 = {f : R → R : f(0) = 1 and f is continuous}. Prove that X1 is an affine subspace of C(R, R) (the space of all continuous function with domain R and mapping to R). Hint: You have to do something with the set X0 = {γ : R → R : f(0) = 0 and f is continuous}.

So, what I think they want me to do at first is two things:
- prove that the set X0 is a subspace of C(R, R) (Closed by addition/multiplication)
- define an arbitrary element β which is an element of C(R, R) but not an element of X0

Once I have done both, I feel like I have to define an arbitrary element of X1, for example, the function ƒ, and show that this function contains both an element of X0 (let's say γ) and the arbitrary element β. In other words:

ƒ(x) = γ(x) + β

If this holds then the set X1 is an Affine Subspace of C(R, R) because it contains the sum of a subspace of C(R, R) and an element of C(R, R)?

I feel like I don't fully understand how to construct this function ƒ(x) and which arbitrary β to define. I have scanned through the answer numerous times and I kinda get their reasoning why, but I can't develop my own reasoning. There must be a logical, stepwise approach to the problem.

Erik

2. Jun 14, 2017

### WWGD

As I understand it, an affine ( and a fine , haha) subspace is just a standard subspace followed by a translation. The subspace just does not go through the origin and there are issues with closure under operations, as in f(0)+g(0)=2 here. EDIT: Like you said, $X0$ is a subspace ( contains the origin) but $X1$ does not.

3. Jun 14, 2017

### Staff: Mentor

I'm not sure what the arbitrariness of $\beta$ has to do here. In general, an affine subspace is simply a linear subspace that got shifted from the origin $\vec{0}$ by a certain, fixed vector $\vec{f}_0\,$.

So you are right, the set $\{f \in C(\mathbb{R},\mathbb{R})\,\vert \,f(0)=0\}$ is the candidate for this subspace and you have to show that it is actually a subspace, i.e. closed under addition and scalar multiplication. (There are other multiplications on $C(\mathbb{R},\mathbb{R})$ so it's better to name them correctly for not to get confused later on.) You also have to make sure that $\vec{0}$ is in this set. This guarantees that this set isn't empty.

In the next step, you have to find this vector $\vec{f}_0\,$, which isn't an arbitrary $\beta$. But you are right as there is more than one possibility in this case. One $\beta = \vec{f}_0\,$ however, will do. What could be a candidate? Chose an easy one, this makes life easier if you want to do anything with this affine space.

4. Jun 14, 2017

### mathwonk

try translating by the constant function 1 or by the function cosine, or by any function at all with value 1 at 0.

5. Jun 14, 2017

### Erik109

Thanks a lot for the fast replies! They are all truly insightful and I surely understand the concept now.
@fresh_42 Special thanks for the extensive explanation.

The only thing is that I struggle a bit with proving a given affine subspace of $C(ℝ, ℝ)$ is indeed an affine subspace. I feel like I have to define 'arbitrary' elements to show that $X_1$ is an affine subspace in general. The answer I was provided started off with:
Fix an arbitrary $f ∈ X_1$. Let $γ: ℝ → ℝ$ be defined by $γ(x) = f(x) + h(x)$, where $h : ℝ → ℝ$ (the vector shifting from the origin, you mentioned) is given by $h ≡ −1$ Note that $γ$ is a continuous function with the property $γ(0) = 0$, i.e. $γ ∈ X_0$.

So, if I understand it correctly, the way they prove it is by showing that an arbitrary element from $X_1$ can be shifted back to $X_0$? So kind of 'deconstructing' an affine subspace to a linear subspace?

PS: Got the hang of the LaTeX structure on the forums, hope this helps with the layout.

6. Jun 14, 2017

### WWGD

Yes, essentially. You can " de-translate" the affine space into a vector space by finding the right amount to translate by.

7. Jun 14, 2017

### Staff: Mentor

Every affine subspace $A$ can be written $A=V +\vec{h}$ with a vector (linear) subspace $V$ and the shifting vector $\vec{h}$.
You can observe two things here:
• $\vec{h}$ is not unique. Every other vector $\vec{h}+\vec{v}$ with $\vec{v}\in V$ does the job as well.
• The difference between two elements $\vec{a}_1 = \vec{v}_1 + \vec{h} \, , \,\vec{a}_2 = \vec{v}_2 + \vec{h}$ in $A$ is $\vec{a}_1 -\vec{a}_2 = (\vec{v}_1 + \vec{h})- (\vec{v}_2 + \vec{h})= \vec{v}_1 - \vec{v}_2$ and always an element of $V$.
Now in your case, $V= \{f_V \in C(\mathbb{R},\mathbb{R})\,\vert \,f_V(0)=0\}$ and $A= \{f_A \in C(\mathbb{R},\mathbb{R})\,\vert \,f_A(0)=1\}$, so you need a function $h=\vec{h} \in C(\mathbb{R},\mathbb{R})$ that moves (shifts) all functions in such a way, that $1=\vec{f}_A(0) = \vec{f}_V(0) +\vec{h}(0) = 0 + h(0)$. @mathwonk mentioned a few candidates for $h$ in post #4. As said above, $\vec{h}$ isn't unique, only up to elements of $V$.

Affine subspaces are not closed under addition nor under scalar multiplication. However, all differences of two vectors are, as they are vectors in the corresponding linear subspace. This way you still have properties like flatness and can calculate similar to what you can do in vector spaces, but you don't have a zero and must be more careful with addition and scalar multiplication. E.g. all tangent spaces $T_p$ are linear spaces, but if you actually want to use them as tangents, you have to consider $\vec{p} + T_p$ which is affine. This is very important as usually it's spoken about the vector space $T_p\,$, e.g. a line $\mathbb{R} \cdot \vec{l}$. And as a line it is a vector space only if it contains the origin $\vec{0}$. The actual tangent are the points $\vec{p} + t\cdot \vec{l}$. I mention this, as it might be confusing if people talk about linearity of tangent spaces, e.g. when talking about a gradient $\nabla$, although it is strictly speaking the affine space shifted by $\vec{p}$.


# Use Pappus' theorem to find the moment of a region limited by a semi-circunference.

This is part of self-study; I found this question in the book "The Calculus with Analytic Geometry" (Leithold).

$R$ is the region limited by the semi-circumference $\sqrt{r^2 - x^2}$ and the x-axis. Use Pappus' theorem to find the moment of $R$ with respect to the line $y = -4$.

Pappus' theorem (also referred to as Guldinus theorem or Pappus-Guldinus theorem) is as follows:

If $R$ is the region limited by the functions $f(x)$ and $g(x)$, then, if $A$ is the area of $R$ and $\bar{y}$ is the y-coordinate of the centroid of $R$, the volume $V$ of the solid of revolution obtained by rotating $R$ around the x-axis is given by: $V = 2\pi\bar{y}A$.

Also, what I'm calling moment (I'm not sure if this is a common term) is the quantity that is divided by the area of the region in order to find a coordinate of the centroid of the region: for example, to find the x-coordinate of the centroid ($\bar{x}$), first one finds the moment around the y-axis ($M_y$), then divides it by the area of the region ($A$).

Since the region $R$ is symmetric with respect to the y-axis, the x-coordinate of the centroid ($\bar{x}$) is zero (therefore, the moment around the y-axis ($M_y$) is zero, because $M_y = \bar{x}A$, where $A$ is the area of the region). So, I only need to find the vertical coordinate of the centroid (with respect to the line $y = -4$) and the moment around the line $y = -4$.

The book's answer for the moment around the line $y = -4$ is: $\frac{1}{2}r^3\left (\pi+\frac{4}{3}\right )$. I included two attempts below; both arrive at a same result, which is different from the result of the book.

Attempt 1

First I tried to use Pappus' theorem to find the vertical coordinate of the centroid of the semi-circular region limited by $\sqrt{r^2 - x^2}$, with respect to the x-axis (not yet with respect to the line y = -4). I will call it $\bar{y}_x$.

Since the solid of revolution obtained by rotating this semi-circular region around the x-axis is a sphere, its volume is $V = \frac{4}{3}\pi r^3$. The area of the semi-circular region is $A=\frac{\pi r^2}{2}$. Substituting $V$ and $A$ into Pappus' theorem:

$\frac{4}{3}\pi r^3 = 2\pi\bar{y}_x\frac{\pi r^2}{2}$

$\bar{y}_x = \frac{4r}{3\pi}$.

This is the vertical coordinate of the centroid with respect to the x-axis. The vertical coordinate of the centroid with respect to the line $y = -4$ is:

$\bar{y} = \frac{4r}{3\pi} + 4$.

To find the moment around the line $y = -4$, I use the fact that $\bar{y} = \frac{M_x}{A}$:

$M_x = \bar{y}A = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$.

Attempt 2

The moment of a plane region with respect to the line $y = -4$ can be found by dividing this region into infinitesimal elements of area, then multiplying the area of each element of area by the vertical coordinate of the centroid.

So, if $f(x) = \sqrt{r^2 - x^2}$, then, if I divide the semi-circular region into several rectangles of length $dx$, the area of each rectangle is $f(x) dx$ and the vertical coordinate of the centroid of each rectangle with respect to the line $y = -4$ is $\frac{f(x)}{2} + 4$. So, the moment with respect to the line $y = -4$ is:

$M_x = \int_{-r}^r \left [ f(x)\times \left ( \frac{f(x)}{2} + 4 \right ) dx \right ]$.

Solving this integral gives the following result:

$M_x = \left ( \frac{4r}{3\pi} + 4\right )\frac{\pi r^2}{2}$,

which is the same result I found in the first attempt.

Both your attempts lead to the correct result, and the solution given in the book is wrong. To see that the book's solution cannot be correct do the following "Gedankenexperiment": Assume that we are told to compute the moment with respect to the line $y=-\eta$ for some given $\eta$ instead of $4$. For an $\eta \gg r$ this moment would be approximatively proportional to $r^2$ and to $\eta$, as is the case in your solution but not in the solution given by the book: The $O(r^3)$ dependency is unwarranted.


# Non-negative polynomials on $[0,1]$ with small integral

Let $P_n$ be the set of degree $n$ polynomials that pass through $(0,1)$ and $(1,1)$ and are non-negative on the interval $[0,1]$ (but may be negative elsewhere).

Let $a_n = \min_{p\in P_n} \int_0^1 p(x)\,\mathrm{d}x$ and let $p_n$ be the polynomial that attains this minimum.

Are $a_n$ or $p_n$ known sequences? Is there some clever way to rephrase this question in terms of linear algebra and a known basis of polynomials?

Following Robert Israel's answer, we also scale everything to $[-1,1]$ (thus multiplying the result by 2). As he mentions, the optimal polynomial is always a square of some other polynomial, $p_{2n}=p_{2n+1}=q_n^2$, and $q_n$ is either even or odd (see Lemma below). So we are left to find the minimal $L_2[-1,1]$-norm of an odd/even polynomial $q_n$ such that $\deg q_n\leq n$ and $q_n(\pm1)=(\pm1)^n$. In other words (recall the division by 2 in the first line!), $a_{2n}=a_{2n+1}=d_n^2/2$, where $d_n$ is the distance from $0$ to the hyperplane defined by $q(1)=1$ in the space of all odd/even polynomials of degree at most $n$ with $L_2[-1,1]$-norm.

Now, this hyperplane is the affine hull of the Legendre polynomials $P_i(x)=\frac1{2^ii!}\frac{d^i}{dx^i}(x^2-1)^i$, where $i$ has the same parity as $n$ (since we know that $P_i(1)=1$ and $P_i(-1)=(-1)^i$). Next, by $\| P_i\|^2=\frac{2}{2i+1}$, our distance is $$\left(\sum_{j\leq n/2}\| P_{n-2j}\|^{-2}\right)^{-1/2} =\left(\sum_{j\leq n/2}\frac{2(n-2j)+1}{2}\right)^{-1/2}=\sqrt{\frac4{(n+1)(n+2)}},$$ attained at $$q_n(x)=\left(\sum_{i\leq n/2}\frac{2}{2(n-2i)+1}\right)^{-1} \sum_{i\leq n/2}\frac{2P_{n-2i}(x)}{2(n-2i)+1}.$$ Thus the answer for the initial question is $a_{2n}=a_{2n+1}=\frac2{(n+1)(n+2)}$ and $p_{2n}=p_{2n+1}=q_n^2$.

Lemma. For every $n$, one of optimal polynomials on $[-1,1]$ is a square of an odd or even polynomial.

Proof. Let $r(x)$ be any polynomial.which is nonnegative on $[-1,1]$ with $r(\pm 1)=1$. We will replace it by some other polynomial which has the required form, has the same (or less) degree and the same values at $\pm 1$, is also nonnegative on $[-1,1]$, and is not worse in the integral sense. Due to the compactness argument, this yields the required result.

Firstly, replacing $r(x)$ by $\frac12(r(x)+r(-x))$, we may assume that $r$ is even (and thus has an even degree $2n$). Let $\pm c_1,\dots,\pm c_{n}$ be all complex roots of $r$ (regarding multiplicities); then $$r(x)=\prod_{j=1}^n\frac{x^2-c_j^2}{1-c_j^2},$$ due to $r(\pm 1)=1$.

Now, for all $c_j\notin[-1,1]$ we simultaneously perform the following procedure.

(a) If $c_j$ is real, then we replace $\pm c_j$ by $\pm x_j=0$. Notice that $$\frac{|x^2-c_j^2|}{|1-c_j^2|}\geq 1\geq \frac{|x^2-0^2|}{|1-0^2|}.$$ for all $x\in[-1,1]$.

(b) If $c_j$ is non-real, then we choose $x_j\in[-1,1]$ such that $\frac{|c_j-1|}{|c_j+1|}=\frac{|x_j-1|}{|x_j+1|}$. Notice that all complex $z$ with $\frac{|c_j-z|}{|x_j-z|}=\frac{|c_j-1|}{|x_j-1|}$ form a circle passing through $-1$ and $1$, and the segment $[-1,1]$ is inside this circle. Therefore, for every $x\in[-1,1]$ we have $$\frac{|c_j-x|}{|x_j-x|}\geq\frac{|c_j-1|}{|x_j-1|},$$ thus $$\frac{|x^2-c_j^2|}{|1-c_j^2|} =\frac{|c_j-x|}{|c_j-1|}\cdot\frac{|c_j+x|}{|c_j+1|} \geq \frac{|x_j-x|}{|x_j-1|}\cdot\frac{|x_j+x|}{|x_j+1|} =\frac{|x^2-x_j^2|}{|1-x_j^2|}.$$ So, we replace $\pm c_j$ and $\pm\bar c_j$ by $\pm x_j$ and $\pm x_j$ (or simply $\pm c_j$ by $\pm x_j$ if $c_j$ is purely imaginary).

After this procedure has been applied, we obtain a new polynomial whose roots are in $[-1,1]$ and have even multiplicities, and its values at $\pm1$ are equal to $1$. So it is a square of some polynomial which is even/odd (since the roots are still split into pairs of opposite numbers). On the other hand, its values at every $x\in[-1,1]$ do not exceed the values of $r$ at the same points, as was showed above. So the obtained polynomial is not worse in the integral sense, as required. The lemma is proved.

• Sorry to be dense, but please say why it has to be the square of another polynomial. I can see that any zeros in the critical interval have even multiplicity, but why can't the polynomial be negative elsewhere (as allowed by OP)? Oct 9 '15 at 22:54
• The last part of the answer (after `finally') is about it; I'll try to expand it. In short: If our polynomial has a non-real root, or if it has real roots outside $[-1,1]$, then we may replace these roots by some root on $[-1,1]$ so as to lower the values of the polynomial at all points in $(-1,1)$(and keep it nonnegative). Oct 10 '15 at 7:05
• Expanded (or, I would say, rewritten). Oct 10 '15 at 7:47

For better symmetry, let me change the boundary points to $(-1,1)$ and $(1,1)$. Then it is clear that we may assume $p_n$ is an even function (and thus you really want $P_n$ to be polynomials of degree at most $n$, otherwise you'll be likely to have an infimum rather than an actual minimum): $p_{2k+1} = p_{2k}$. I get \eqalign{p_0(x) = 1, & a_0 = 2\cr p_2(x) = x^2, & a_2 = \dfrac{2}{3}\cr p_4(x) = \dfrac{(5 x^2 - 1)^2}{16},& a_4 = \dfrac{1}{3}\cr p_6(x) = \dfrac{x^2}{16} (7 x^2 - 3)^2,& a_6 = \dfrac{1}{5}\cr p_8(x) = \dfrac{1}{64} (21 x^4 - 14 x^2 + 1)^2, & a_8 = \dfrac{2}{15}\cr p_{10}(x) = \dfrac{x^2}{64} (33 x^4 - 30 x^2 + 5)^2, & a_{10} = \dfrac{2}{21}\cr p_{12}(x) = \dfrac{1}{4096} (429 x^6 - 495 x^4 + 135 x^2 - 5)^2, & a_{12} = \dfrac{1}{14}} Well, so far it looks like $a_{2n} = \dfrac{4}{(n+1)(n+2)}$

• How were these computed? Oct 9 '15 at 0:03
• The leading coefficients look like Catalan numbers when the polynomials are normalized to $\sqrt{2^{2n} p_{2n}(x)}$. Oct 9 '15 at 1:51
• The next coefficients match $-{2n \choose n-2}$. The whole triangle looks like oeis.org/A234950 read skewed, with alternating signs. See oeis.org/A062991 for a signed version. Oct 9 '15 at 3:43
• For $n$ even, $$p_{2n} = \prod_{i=1}^{n/2} \dfrac{(x^2 -c_i^2)^2}{(1-c_i^2)^2}$$ For $n$ odd, there's also a factor of $x^2$. Differentiate $a_{2n} =\int_{-1}^1 p_{2n}(x)\; dx$ with respect to each $c_i$, solve, check which solution gives the least $a_{2n}$. Oct 9 '15 at 4:49


Question

# Assertion :The chord of contact of tangent from a point $$P$$ to a circle passes through $$Q$$. If $${ l }_{ 1 }$$ and $${ l }_{ 2 }$$ are the lengths of the tangents from $$P$$ and $$Q$$ to the circle, then $$PQ$$ is equal to $$\sqrt { { { l }_{ 1 } }^{ 2 }+{ { l }_{ 2 } }^{ 2 } }$$ Reason: The equation of chord of contact of tangents from the point $$P\left( { x }_{ 1 },{ y }_{ 1 } \right)$$ to the circle $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$ is $$x{ x }_{ 1 }+y{ y }_{ 1 }={ a }^{ 2 }$$

A
Assertion is true, Reason is true and reason is correct explanation for Statement-1.
B
Assertion is true, Reason is true and Reason is NOT correct explanation for Assertion.
C
Assertion is true, Reason is false
D
Assertion is false, Reason is true

Solution

## The correct option is A Assertion is true, Reason is true and reason is correct explanation for Statement-1.Let $$P\equiv \left( { x }_{ 1 },{ y }_{ 1 } \right)$$ and $$Q\equiv \left( { x }_{ 2 },{ y }_{ 2 } \right)$$Let the equation of the given circle be $${ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$The equation of chord of contact of tangents from the point $$P\left( { x }_{ 1 },{ y }_{ 1 } \right)$$ to the given circle is $$x{ x }_{ 1 }+y{ y }_{ 1 }={ a }^{ 2 }$$Since it passes through $$Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$$$$\therefore { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 }={ a }^{ 2 }$$    ...(1)Now $${ l }_{ 1 }=\sqrt { { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }-{ a }^{ 2 } } ,{ l }_{ 2 }=\sqrt { { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 }-{ a }^{ 2 } }$$and $$PQ=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } }$$$$=\sqrt { \left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 } \right) +\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 } \right) -2\left( { x }_{ 1 }{ x }_{ 2 }+{ y }_{ 1 }{ y }_{ 2 } \right) } \\ =\sqrt { \left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 } \right) +\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 } \right) -2{ a }^{ 2 } } \\ =\sqrt { \left( { { x }_{ 1 } }^{ 2 }+{ { y }_{ 1 } }^{ 2 }-{ a }^{ 2 } \right) +\left( { { x }_{ 2 } }^{ 2 }+{ { y }_{ 2 } }^{ 2 }-{ a }^{ 2 } \right) } =\sqrt { { { l }_{ 1 } }^{ 2 }+{ { l }_{ 2 } }^{ 2 } }$$Maths

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# Find the smallest $n \in \mathbb{N}$ such that the group is isomorphic to the direct product of $n$ cyclic groups

Find the smallest $$n \in \mathbb{N}$$ such that the group $$\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$$ is isomorphic to the direct product of $$n$$ cyclic groups.

I'm not sure but if I understand correctly,

$$\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$$ is isomorphic to $$\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{4} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5} \times \mathbb{Z}_{9}$$,

$$\mathbb{Z}_{2} \times \mathbb{Z}_{5} \times \mathbb{Z}_{9}$$ is isomorphic to $$\mathbb{Z}_{90}$$,

$$\mathbb{Z}_{3} \times \mathbb{Z}_{4} \times \mathbb{Z}_{5}$$ is isomorphic to $$\mathbb{Z}_{60}$$,

and therefore, $$\mathbb{Z}_{6} \times \mathbb{Z}_{20} \times \mathbb{Z}_{45}$$ is isomorphic to $$\mathbb{Z}_{60} \times \mathbb{Z}_{90}$$ and the answer is $$n = 2$$. Is this a correct solution?

• I agree that the answer is $n=2$. For the sake of being thorough, I would probably explain why $n=1$ is impossible (just because you've found one situation where $n=2$ doesn't necessarily mean it's the smallest $n$). – Theo C. Apr 23 at 21:21
• Yes, it is. You just have to find the smallest number of groups of moduli such that the moduli in each group are pairwise coprime. – Bernard Apr 23 at 21:24
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun May 24 at 14:27

You may find the invariant factors.

Actually $$\mathbb{Z}_6\times \mathbb{Z}_{20}\times \mathbb{Z}_{45}\cong (\mathbb{Z}_2\times\mathbb{Z}_4)\times(\mathbb{Z}_3\times\mathbb{Z}_9)\times(\mathbb{Z}_5\times\mathbb{Z}_5)$$. Now we can pick the largest ones in each bracket to form $$\mathbb{Z}_4\times\mathbb{Z}_9\times\mathbb{Z}_5\cong\mathbb{Z}_{180}$$. Then we pick the largest ones of the remaining, which is $$\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_5\cong\mathbb{Z}_{30}$$, and nothing remain. Hence the group is isomorphic to $$\mathbb{Z}_{180}\times\mathbb{Z}_{30}$$ and so $$n\le 2$$. Since the group is obviously not cyclic, we have $$n = 2$$.

I think we can always get $$n$$ for any abelian group by finding the invariant factors.

According to GAP, the group is $$\Bbb Z_{180}\times \Bbb Z_{30}$$ and, of course, not cyclic.

gap> G:=DirectProduct(CyclicGroup(6) , DirectProduct(CyclicGroup(20), CyclicGroup(45)));
<pc group of size 5400 with 8 generators>
gap> StructureDescription(G);
"C180 x C30"
gap> IsCyclic(G);
false
gap>


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