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# Multivariable Calculus and Matrix Algebra (Herron): Week 5 Notes (2/22/16 - 2/26/16) MATH 2010

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This 5 page Class Notes was uploaded by creask on Saturday February 27, 2016. The Class Notes belongs to MATH 2010 at Rensselaer Polytechnic Institute taught by Isom Herron in Spring 2016. Since its upload, it has received 15 views. For similar materials see Multivariable calculus and matrix algebra in Mathematics (M) at Rensselaer Polytechnic Institute.

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## Reviews for Multivariable Calculus and Matrix Algebra (Herron): Week 5 Notes (2/22/16 - 2/26/16)

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Date Created: 02/27/16

MATH 2010 - Multivariable Calculus & Matrix Algebra Professor Herron - Rensselaer Polytechnic Institute Week 4 (2/22/16 - 2/26/16) Important: These notes are in no way intended to replace attendance in lecture. For best results in this course, it is imperative that you attend lecture and take your own detailed notes. Please keep in mind that these notes are written speci▯cally with Professor Herron’ s sections in mind, and no one else’ s. Double Integrals ZZ XX n f(x;y)dA = lim f(x ;y )▯A m;n!1 ij ij ij R i=1 j=1 If f(x;y) ▯ D, the integral is the volume that lies above a rectangle and below the surface f. Example: Midpoint Rule for Double Integrals (x +x ) (j▯1+yj) RR P Pn let▯i= i21 i and y▯j= 2 . Then: R f(x;y)dA ▯ f(▯i;▯j)▯A ij i=1 j=1 RRe the midpoint rule for m = n = 2 to estimate the value of the double integral (x + y ▯ 2)dA, where R = ((x;y)j1 ▯ x ▯ 4;1 ▯ y ▯ 3). R 7 13 3 5 By the midpoint rule, ▯1= ;4 ▯2= 4;y1 = ;2 ▯2= .2 5 3 Thus, the area of each sub-rectangle is ▯A = 1 ▯ 2 ▯ 1) = 2 Subbing x and y values back into the double integral: RR ▯5 ▯3 ▯9▯ 3 ▯11 3 ▯ 15 3 { R(x + y ▯ 2)dA ▯ 4 2 + 4 2+ 4 2 + 4 2 = 15 { x + y + z = 2 Double Integrals for a Constant Function Suppose f(x;y) = k. ZZ ZZ n m XX f(x;y)dA = k dA = lim k▯A ij R R n;m!1 i=1 j=1 Xn Xm = k n;m!1 ▯A =ijA(R) i=1 j=1 1 A(R) is the area of R. RR A(R) = (b ▯ a)(d ▯ c). Thus: Rk dA = k(b ▯ a)(d ▯ c) Use of Iterated Integrals to Find Double Integrals ZZ Z Z Z d b d f(x;y)dA = f(x;y)dxdy = S(y)dy = V R c a c Fubini’s Theorem: { If f is continuous on rectangle R = ((x;y)ja ▯ x ▯ b;c ▯ y ▯ d), then: RR R R R Rb { f(x;y)dA = f(x;y)dy dx = f(x;y)dxdy R a c c a { From the midpoint rule example: R4 ▯1 ▯4 * S(y) = (x + y ▯ 2)dx = 2x + xy + 2x 1 1 1 1 * = (26) + 4y ▯ 8 ▯ ( + 2 ▯ 2) 1 3 * = 8▯+ 4y ▯ 8▯▯ 2▯ y + 2 ! 3y + 2 R3 R3 ▯ ▯ * S(y)dy = (3y + )dy = 3y + y 3 3 2 2 2 1 1 1 * = 36▯ 6 = 30= 15 2 2 2 * Integrating dy ▯rst yields the same result. Example: Iterated Integrals R R Calculate 5 3(x + y )dxdy. 2 1 R ▯ ▯3 R5▯ ▯ R ▯ ▯ = 1x + xy 2 = 9 + 3y ▯ 1▯ y 2 dy = 26 + 2y 2 2 3 1 2 3 2 3 ▯26 2 3▯5 130 250 52 16 312 = 3 y + 3 2 = 3 + 3 ▯ 3 ▯ 3 = 3 = 104 Three Important Properties RR RR 1. If f(x;y) ▯ g(x;y) for all (x;y) in R, then R f(x;y)dA ▯ R g(x;y)dA. RR RR RR 2. R [f(x;y) + g(x;y)]dA = R f(x;y)dA + R g(x;y)dA. RR RR 3. cf(x;y)dA = c f(x;y)dA, where c is a constant. R R Equation 5 (Iterated Integrals) and Examples ZZ Z b Z d g(x)h(y)dA = g(x)dx h(y)dy;R = [a;b] ▯ [c;d] R a c 2 5 3 R R 2 2 Example 1: x y dxdy 2 1 ▯ ▯▯ ▯ R5 R3 ▯ ▯3 ▯ ▯5 { = y dy x dx = 1x3 1y 3 2 1 3 1 3 2 26 117 { = 3 ▯ 3 = 338 Example 2 RR { The double integral of the constant function f(x;y) = k has the property k dA = RR R ▯ A(R▯, where A(R) is the area of R. Find f dA for f = 2 and R = [▯1;1] ▯ 1;3 RR2 2 { f dA = f A(R) by the above property. { We know f = 2 and we can solve for the area by multiplying the length and width ▯3 1▯ given by the domain of R: A(R) = (1 ▯ (▯1)) ▯ 2 ▯ 2 = 2 ▯ 1 = 2: RR { Thus, f dA = 2 ▯ 2 = 4. R Example 3: Find f(x;y)dx if f(x;y) = 8x + 6x y.2 0 { Notice that there is only one integration we need to take: that with respect to x. R2 { (8x + 6x y) = [4x + 2x y] = 16+16y. 0 0 More General Regions De▯ne F with domain R by the following: ( F(x;y) = f(x;y)if(x;y)is inD 0iff(x;y)is inRbut not inD If F is integrable over R, de▯ne the double integral of f over D by: ZZ ZZ f(x;y)dA = F(x;y)dA D R If f is continuous on a type I region D such that: { D = f(x;y)ja ▯ x ▯ b;g (x)1▯ y ▯ g (x)g2(vertically simple), then: b g2(x) RR R R { D f(x;y)dA = f(x;y)dy dx. ag1(x) { This means that D has a "top" and a "bottom" boundary curve. If D = D [1D , 2here D and 1 don’t 2verlap except, perhaps, on their boundaries, then: ZZ ZZ ZZ f(x;y)dA = f(x;y)dA = f(x;y)dA D D1 D 2 3 Type II Regions (Horizontally Simple) h (y) RR R R2 { D f(x;y)dA = f(x;y)dxdy, where D is a type II region. c h1(y) Example: Rearranging the Order of Integration R R1 Given the double integral f(x;y)dy dx, change the order of integration. 0 x2 p p R R y y = x , so then x = y. Then the new order is: f(x;y)dy dx. 0 0 Regions can also be neither type I or type II, but can be expressed as a union of regions of type I or type II. Example: Double Integral of a Constant Function RR The double integral of the constant function f(x;y) = k has the property k dA = D k A(D) for region D, where A(D) is equal to the area of D. RR Find f dA when f = 2 and D is a disk with center at the origin and radius of 2. D { From the property above, f ▯ dA = f ▯ A(D). { Thus, (2) ▯ (2 ▯) = 8▯. Average Value of a Function on a Region D RR ▯ D f dA f = A(D) Tra▯c science: ▯nd the average distance of points in a disk of radius a from the origin. RR p 2 2 { d = N x +y dA ▯a2 { It is di▯cult to solve this in the Cartesian coordinate system, so convert the equation to polar coordinates. RR p 2 Z 2▯Z a ▯ ▯a ▯ N r r dr d▯ 1 2 1 2▯ 1 3 2a d = 2 = 2 r dr d▯ = 2[▯]0 r = ▯a ▯a 0 0 ▯a 3 0 3 This means that there are "more points near the perimeter" (ring roads). Triple Integrals Useful for centroids, the "average" coordinates in three dimensions. RRR x dV RRR y dV RRR z dV ▯ = W ,▯ = W , ▯ = W V V V 4 Other Coordinate Systems Cylindrical: x = r cos▯, y = r sin▯, z = z Spherical End of Document 5

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