w7-C - Math 20C Multivariable Calculus Lecture 19 1 ' $...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 20C Multivariable Calculus Lecture 19 1 ' $ Double integrals (Sec. 15.1 - 15.2) Review of the integral of single variable functions. Definition of a double integral on rectangles. Average of a function. Examples of double integrals in rectangles (sec. 15.2) Slide 1 & % ' Integral of a single variable function Definition 1 Let f (x) be a function defined on a interval x [a, b]. The integral of f (x) in [a, b] is the number given by b n $ Slide 2 a f (x)dx = lim n f (x ) x, i i=0 & if the limit exists. Given a natural number n we have introduced a partition on [a, b] given by x = (b - a)/n. We denoted x = (xi + xi-1 )/2, where xi = a + ix, i i = 0, 1, , n. This choice of the sample point x is i called midpoint rule. % Slide 3 Slide 4 Math 20C Multivariable Calculus & & y ' ' if the limit exists. x = (x1 - x0 )/n, y = (y1 - y0 )/n. Let x = (xi + xi-1 )/2, i Given a natural number n, the partition on R are rectangles of side i, j = 0 , n. These sample points x , yj are called midpoint rule. i yj = (yj + yj-1 )/2, where xi = x0 + ix, and yj = y0 + jy, for Definition 2 Let f (x, y) be a function defined on a rectangle R = [x0 , x1 ] [y0 , y1 ]. The integral of f (x, y) in R is the number given by Double integrals on rectangles Riemann sum of a single variable function R x0 f (x) dxdy = lim x1 n x2 i=0 j=0 n f(x) n x3 f (x , yj ) xy, i Lecture 19 x4 y % $ % $ 2 Math 20C Multivariable Calculus Lecture 19 3 ' Partition of the domain of a two variable function $ z Slide 5 y0 x3 x2 x1 x0 y1 y2 y3 y 4 y x4 x & % ' Double integrals of f (x, y) are volumes in I 3 R If f (x, y) 0, then f (x, y) dxdy = V the volume R above R and below the surface given by the graph of f (x, y). z f(x,y) $ Slide 6 y R x & % Math 20C Multivariable Calculus Lecture 19 4 ' The order of integration can be switched in double integrals of continuous functions Theorem 1 (Fubini) If f (x, y) is a continuous function in R = [x0 , x1 ] [y0 , y1 ], then y1 x1 $ Slide 7 f (x, y) dxdy = R y0 x1 x0 y1 f (x, y) dx dy, f (x, y) dy dx. x0 y0 = & Notation: One also denotes the double integral as Z y1 Z x 1 Z Z f (x, y) dxdy. f (x, y) dxdy = R y0 x0 % ' Here is an example of a double integral Z 3 1 $ = Slide 8 1 2 " 2 2 " 2 3 " 3 2 " y x 0 + y x 0 dy, 2 3 1 Z 3 16 3 y dy, = 2y 2 + 3 1 3 3 2 3 + 16 y 4 , = y 3 1 12 1 2 4 = 26 + 80. 3 3 = Z 3 Z Z 1 2 0 3 (xy 2 + 2x2 y 3 )dxdy = 2 Z 0 (xy 2 + 2x2 y 3 )dx dy, & % Math 20C Multivariable Calculus Lecture 19 5 ' Second example Z 4 1 $ Z 2 1 ,, y x + y x dydx = Slide 9 y x + dy dx, y x 1 1 Z 4 ` 1 " 2 2 " = y 1 dx, x ln(y)|2 + 1 2x 1 Z 4 3 = dx, ln(2)x + 2x 1 1 4 3 = ln(2) x2 1 + ln(x)|4 , 1 2 2 3 15 ln(2) + ln(4), = 2 2 ,, 15 + 3 ln(2). = 2 Z 4 2 Z ,, & % ' Fubini theorem in the case of f (x, y) = g(x)h(y): $ x1 x0 y1 x1 y1 g(x)h(y)dydx = y0 x0 g(x)dx y0 h(y)dy . Slide 10 Example: 2 0 0 1 1 + x2 dydx = 1 + y2 = 2 1 (1 + x )dx 0 0 2 1 dy , 1 + y2 1 x|2 + x|2 0 3 0 8 = 2+ . 4 3 & arctan(y)|1 , 0 % Slide 11 Math 20C Multivariable Calculus ' The number f given by f = average of f (x) in [a, b]. Recall the average of f (x) in [a, b] f Slide 12 & ' & is the average of a function f (x, y) in the domain R = [x0 , x1 ] [y0 , y1 ], where Definition 3 (Average) The number f given by the area of the rectangle domain R. The average of f (x, y) in R y f= A(R) = (x1 - x0 )(y1 - y0 ) a 1 A(R) R 1 b-a f (x, y) dxdy, a b f (x) dx is the Lecture 20 f(x) b x % $ % $ 6 Math 20C Multivariable Calculus Lecture 20 7 ' $ Double integrals on regions Regions in Cartesian coordinates (Sec. 15.3) Type I: Regions functions y(x). Slide 13 Type II: Regions functions x(y). Regions in Cartesian coordinates (Sec. 15.4) Type I: Regions functions r(). Type II: Regions functions (r). & % ' $ Regions in Cartesian coordinates y(x): Type I z f(x,y) Slide 14 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 8 ' $ Regions in Cartesian coordinates y(x): Type I Theorem 2 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 15 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 16 g(x) = x 1 g(x) = x 2 0 0 & 1 x % Math 20C Multivariable Calculus Lecture 20 9 ' 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 17 = 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 18 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 10 ' $ Regions in Cartesian coordinates x(y): Type II Theorem 3 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 19 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 20 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 11 ' Z Z Z 1 0 $ Z y f (x, y) dxdy = D Slide 21 " " " " 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, & % ' $ 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 22 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 20 12 ' $ Review of polar coordinates Definition 4 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 r= x2 + y 2 , = arctan y . x Slide 23 The inverse expression is x = r cos(), & y = r sin(). % ' $ Double integrals in polar coordinates on disk sections Theorem 4 If f (r, ) in continuous in Slide 24 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 Lecture 20 13 ' $ 1 x2 + y 2 2} Compute the integral of f (x, y) = x2 + 2y 2 on D = {(x, y) I 2 : 0 y, R 0 x, Translate to polar coordinates. x = r cos(), y = r sin(). Then f (r, ) = r 2 + r2 sin2 (). Slide 25 The region D is D = (r, ) I 2 : 0 R Z Z f (r, )dA = D , 2 1r 2 . = "Z Z /2 0 /2 Z 2 1 (1 + sin ()) d 0 r2 (1 + sin2 ())r drd, # "Z 2 2 3 1 r dr , # & % ' $ Example: Continuation Z Z # 1 4 2 1 (1 - cos(2)) d (r |1 ) , + 2 4 0 1 /2 1 3 /2 + (|0 ) - (sin(2)|0 ) , 2 2 4 4 3 h i + , 4 2 4 9 . 16 " /2 (|0 ) f (r, )dA D = = = = Slide 26 Z /2 & % Math 20C Multivariable Calculus Lecture 20 14 ' Integrate f (x, y) = e-(x D = {(r, ) R2 : 0 , 2 2 +y 2 ) $ on 0 r 2} Slide 27 Notice, f (r, ) = e-r , then, Z Z Z 2 2 e-(x +y ) dA = D 2 0 Z 2 e-r r drd, 2 0 substitute u = r , 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 & % ' $ Summarizing, from Cartesian to polar Theorem 5 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 28 f (r cos(), r sin())r drd. & % Math 20C Multivariable Calculus Lecture 20 15 ' $ Type I in polar coordinates Theorem 6 Let 0 < h0 () h1 () be two continuous functions defined on an interval [0 , 1 ]. Let f (r, ) be a continuous function in Slide 29 D = {(r, ) I 2 : 0 < h0 () r h1 (), R 0 1 }. 1 h1 () Then, the integral of f (r, ) in D is given by f (r, ) dA = D 0 h0 () f (r, )r dr d. % & ' Type II in polar coordinates $ Theorem 7 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 30 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. % & ...
View Full Document

Ask a homework question - tutors are online