w8-C - Math 20C Multivariable Calculus Lecture 20 1 ' $...

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Unformatted text preview: Math 20C Multivariable Calculus Lecture 20 1 ' $ Double integrals on regions Slide 1 Regions in Cartesian coordinates (Sec. 15.3) Type I: Regions functions y(x). Type II: Regions functions x(y). & % ' $ Regions in Cartesian coordinates y(x): Type I z f(x,y) Slide 2 y g(x) 1 f(x,g(x)) 0 f(x,g(x)) 1 x0 y g(x) 0 x1 g(x) 0 g(x) 1 x0 x1 x x & % Math 20C Multivariable Calculus Lecture 20 2 ' $ Regions in Cartesian coordinates y(x): Type I Theorem 1 Let g0 (x), g1 (x) be two continuous functions defined on an interval [x0 , x1 ], and such that g0 (x) g1 (x). Let f (x, y) be a continuous function in D = {(x, y) I 2 : x0 x x1 , R x1 Slide 3 g0 (x) y g1 (x)}. g1 (x) Then, the integral of f (x, y) in D is given by f (x, y) dxdy = D x0 g0 (x) f (x, y)dy dx. & % ' $ Cartesian Type I: Find the f (x, y) = x2 + y 2 , on . y D f (x, y) dxdy for x2 y x} D = {(x, y) I 2 : 0 x 1, R Slide 4 g(x) = x 1 g(x) = x 2 0 0 & 1 x % Math 20C Multivariable Calculus Lecture 20 3 ' 1 x x2 1 $ f (x, y) dxdy = D 0 (x2 + y 2 )dy dx, 1 3x y |x 2 3 dx, = 0 1 x2 (y|x2 ) + x Slide 5 = 0 1 1 x2 (x - x2 ) + (x3 - x6 ) dx, 3 1 1 x3 - x4 + x3 - x6 dx, 3 3 1 = 0 = 1 4 1 5 1 1 x - x + x4 - x7 4 5 12 21 1 1 1 9 = - - = . 3 5 21 357 , 0 & % ' $ Regions in Cartesian coordinates x(y): Type II y z f(x,y) Slide 6 y1 f(h(y),y) 0 f(h(y),y) 1 y0 y1 h(y) 0 y y h(y) 0 h(y) 1 0 h(y) 1 x x & % Math 20C Multivariable Calculus Lecture 20 4 ' $ Regions in Cartesian coordinates x(y): Type II Theorem 2 Let h0 (y), h1 (y) be two continuous functions defined on an interval [y0 , y1 ], and such that h0 (y) h1 (y). Let f (x, y) be a continuous function in D = {(x, y) I 2 : h0 (y) x h1 (y), R y1 h1 (y) Slide 7 y0 y y1 }. Then, the integral of f (x, y) in D is given by f (x, y) dxdy = D y0 h0 (y) f (x, y)dx dy. & % ' Cartesian Type II: Find the D f (x, y) dxdy for x2 y x} $ f (x, y) = x2 + y 2 , on . y 1 D = {(x, y) I 2 : 0 x 1, R Slide 8 h(y)=y 0 h(y) = y 1 0 1 x Notice that h0 (y) = y, and h1 (y) = y. Then, y, y0 y y1 }. D = {(x, y) I 2 : h0 (y) = y x h1 (y) = R & % Math 20C Multivariable Calculus Lecture 20 5 ' Z Z Z 1 0 $ Z y f (x, y) dxdy = D Slide 9 " " " " 1 y y x3 |y = + y 2 x|y dy, 3 0 Z 1 1 3/2 (y - y 3 ) + y 2 (y 1/2 - y) dy, = 3 0 Z 1 1 3/2 1 3 y - y + y 5/2 - y 3 dy, = 3 3 0 1 1 2 5/2 1 1 4 2 7/2 1 4 = y - y + y - y , 35 34 7 4 0 2 1 2 1 9 = - + - = . 15 12 7 4 357 Z 1 y (x2 + y 2 )dx dy, & % ' $ Integrate f (x, y) = 2xy in the region bounded by y = 0, y = x and y + x = 2 y 2 y = -x + 2 y=x 1 D 1 1 x Slide 10 2 2 x 2-x f dxdy = D 0 0 2xy dy dx + 1 0 2xy dy dx. % & Math 20C Multivariable Calculus Lecture 20 6 ' $ Integrate f (x, y) = 2xy in the region bounded by y = 0, y = x and y + x = 2 y 2 y = -x + 2 y=x 1 D 1 1 Slide 11 2 2-y x f dxdy = D 0 y 2xy dx dy. % & ' $ Find the n D 2 f (x, y) dxdy for f (x, y) = 1, and x2 9 D = (x, y) I : R + y2 4 1 o Slide 12 As type I, then, g1 (x) = 3 1 - y 2 /4, As type II, then, h1 (x) = 2 1 - x2 /9, & g0 (x) = -3 h0 (y) = -2 1 - y 2 /4. 1 - x2 /9. % Math 20C Multivariable Calculus Lecture 21 7 ' $ D 2 Find the n f (x, y) dxdy for f (x, y) = 1, and x2 9 D = (x, y) I : R + y2 4 1 o y 2 x /4 + y /9 = 1 2 2 3 x Slide 13 y 2 2 y = g 1(x) = 1- x /9 y 2 2 x = h (y) = 1- y /4 1 3 2 x 2 x = h (y) = - 1- y /4 0 3 x y = g (x) = - 1- x /9 0 & % ' $ Double integrals on regions in polar coordinates Slide 14 Regions in Cartesian coordinates (Sec. 15.4) Type I: Regions functions r(). Type II: Regions functions (r). & % Math 20C Multivariable Calculus Lecture 21 8 ' Review of polar coordinates $ Slide 15 Definition 1 Let (x, y) be Cartesian coordinates in I 2 . R Then, polar coordinates (r, ) are defined in I 2 - {(0, 0)}, and given by R y . r = x2 + y 2 , = arctan x The inverse expression is x = r cos(), y = r sin(). & y r 0 x % ' $ Double integrals in polar coordinates on disk sections Theorem 3 If f (r, ) in continuous in Slide 16 D = {(r, ) : 0 < r0 r r1 , 1 r1 0 1 < 2}, f (r, ) r drd. then D f (r, ) dA = 0 r0 Disk sections in polar coordinates rectangular sections in Cartesian coordinates & % Math 20C Multivariable Calculus ' Disk sections in polar coordinates rectangular sections in Cartesian coordinates y y y Slide 18 Slide 17 & & ' The region D is D = (r, ) I 2 : 0 R Compute the integral of f (x, y) = x2 + 2y 2 on Translate to polar coordinates. x = r cos(), y = r sin(). Then D = {(x, y) I 2 : 0 y, R y0 1 x0 x0 x x 1 , Z Z y0 y y 1 , D f (r, )dA = f (r, ) = r 2 + r2 sin2 (). = x1 "Z Z 0 0 x, /2 0 /2 x Z (1 + sin ()) d 1 2 1 x2 + y 2 2} 0 0 1 2. r2 (1 + sin2 ())r drd, # "Z 2 0 r0 r r1 , r0 , 2 01 Lecture 21 1r 00 1 2 r dr , 2 . 3 r1 # x % $ % $ 9 Math 20C Multivariable Calculus Lecture 21 10 ' $ Example: Continuation Z Z # 1 1 4 2 f (r, )dA = + (1 - cos(2)) d (r |1 ) , 2 4 D 0 1 /2 1 3 /2 = + (|0 ) - (sin(2)|0 ) , 2 2 4 4 i 3 h , + = 4 2 4 Z Z 9 9 = f (r, )dA = , . 16 16 D /2 (|0 ) " Slide 19 Z /2 & % ' Integrate f (x, y) = e-(x D = {(r, ) R : 2 2 +y 2 ) on Z Z $ 0 , 2 0 r 2} e-(x 2 +y 2 ) Notice, f (r, ) = e-r , then, Slide 20 substitute u = r 2 , then du = 2r dr, then Z Z Z Z 2 2 1 4 -u e-(x +y ) dA = e dud 2 0 0 D Z 1 (-e-u |4 )d, = 0 2 0 ,, 1 = 1- 4 , 2 e Z Z e-(x D 2 Z Z 2 0 dA = e-r r drd, 2 D 0 +y 2 ) dA = 2 ,, 1 1- 4 . e & % Math 20C Multivariable Calculus Lecture 21 11 ' $ Summarizing, from Cartesian to polar Theorem 4 Let f (x, y) be a continuous function on a domain D, where (x, y) represent Cartesian coordinates. Let (r, ) be polar coordinates. Then the following formula holds, f (x, y) dxdy = D D Slide 21 f (r cos(), r sin())r drd. & % ' Domains in type I in polar coordinates ( (r, ) I 2 : R 0 h0 () r h1 (), 0 1 ) $ D= . Slide 22 y h1(0) h0(0) 00 x % 01 & Math 20C Multivariable Calculus Lecture 21 12 ' $ Type I in polar coordinates Theorem 5 Let 0 < h0 () h1 () be two continuous functions defined on an interval [0 , 1 ]. Let f (r, ) be a continuous function in Slide 23 D= ( (r, ) I : R 2 0 h0 () r h1 (), 0 1 ) . Then, the integral of f (r, ) in D is given by 1 h1 () f (r, ) dA = D 0 h0 () f (r, )r dr d. & % ' Find the area of intersection of the two circles r = cos() and r = sin() y $ r=sin(0) Slide 24 1/2 r=cos(0) 1/2 x & % Math 20C Multivariable Calculus Lecture 21 13 ' Find the area of intersection of the two circles r = cos() and r = sin() A= = Z /4 0 /4 $ Slide 25 1 cos2 () d, /4 2 0 Z /4 Z /2 1 1 = [1 - cos(2)] d + [1 + cos(2)] d, 4 0 /4 4 " " 1 " " 1 1 /4 /2 - 0 - sin(2)|0 + - + sin(2)|/4 , = 4 4 2 2 4 2 ,, ,, 1 1 1 - -0 + 0- , = 4 2 2 2 1 sin2 () d + 2 = 1 - 8 4 A= 1 ( - 2) . 8 Z Z sin() r dr d + 0 Z /2 Z /2 /4 Z cos() r dr d, 0 & % ' Find the volume between the sphere x2 + y 2 + z 2 = 1 and the cone z = x2 + y 2 z $ z Slide 26 R x y y x z= & 1 - r2, z = r. % Math 20C Multivariable Calculus Lecture 21 14 ' $ Find the volume between the sphere x2 + y 2 + z 2 = 1 and the cone z = x2 + y 2 z z z Slide 27 y x = y x - x y 2 r0 0 V = 0 2 r0 1 - r 2 (rdr)d - r (rdr)d. 0 0 & % ' Find the volume between the sphere x2 + y 2 + z 2 = 1 and the cone z = x2 + y 2 ,,Z 2 $ V = d 0 1/ 0 = 2 Slide 28 = 2 Z 1 1 1 u1/2 du - r3 2 1/2 3 0 1 1 1 2 3/2 1 = 2 u - , 2 3 1/2 3 23/2 1 1 2 1 - 3/2 - 3/2 , = 3 2 2 Z 2 Z p 1/ 0 2 hp 1- r2 - r rdr 2 1 - r2 rdr - 0 1/2 Z 1/ r2 dr , i ! , u = 1 - r2 , V = " " 2- 2 . 3 & % Math 20C Multivariable Calculus Lecture 21 15 ' Type II in polar coordinates $ Theorem 6 Let g0 (r), g1 (r) be two continuous functions defined on an interval [r0 , r1 ], and such that 0 < g0 (r) g1 (r) < 2. Let f (r, ) be a continuous function in Slide 29 D = {(r, ) I 2 : 0 < r0 r r1 , R Then, the integral of f (r, ) in D is given by r1 g1 (r) 0 < g0 (r) g1 (r) < 2}. f (r, ) dA = D r0 g0 (r) f (r, )d rdr. % & ...
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This note was uploaded on 04/30/2008 for the course MATH 20C taught by Professor Helton during the Spring '08 term at UCSD.

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