w8-C - Math 20C Multivariable Calculus Lecture 21 1 Slide 1...

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Unformatted text preview: Math 20C Multivariable Calculus Lecture 21 1 Slide 1 ’ & \$ % Double integrals on regions (Sec. 15.3) • Regions function of y . • Regions function of x . • Properties of double integrals. Slide 2 ’ & \$ % Regions functions of y Theorem 1 (Type I) Let g ( x ) , g 1 ( x ) be two continuous functions defined on an interval [ x , x 1 ] , and such that g ( x ) ≤ g 1 ( x ) . Let f ( x, y ) be a continuous function in D = { ( x, y ) ∈ IR 2 : x ≤ x ≤ x 1 , g ( x ) ≤ y ≤ g 1 ( x ) } . Then, the integral of f ( x, y ) in D is given by Z Z D f ( x, y ) dxdy = Z x 1 x " Z g 1 ( x ) g ( x ) f ( x, y ) dy # dx. Math 20C Multivariable Calculus Lecture 21 2 Slide 3 ’ & \$ % Example: Type I • Find the R R D f ( x, y ) dxdy for f ( x, y ) = x 2 + y 2 , D = { ( x, y ) ∈ IR 2 : 0 ≤ x ≤ 1 , x 2 ≤ y ≤ x } . Slide 4 ’ & \$ % Z Z D f ( x, y ) dxdy = Z 1 ·Z x x 2 ( x 2 + y 2 ) dy ¸ dx, = Z 1 · x 2 ( y | x x 2 ) + 1 3 ( y 3 | x x 2 ) ¸ dx, = Z 1 · x 2 ( x- x 2 ) + 1 3 ( x 3- x 6 ) ¸ dx, = Z 1 · x 3- x 4 + 1 3 x 3- 1 3 x 6 ¸ dx, = 1 4 x 4 | 1- 1 5 x 5 | 1 + 1 12 x 4 | 1- 1 21 x 7 | 1 , = 1 3- 1 5- 1 21 = 9 3 × 5 × 7 . Math 20C Multivariable Calculus Lecture 21 3 Slide 5 ’ & \$ % Regions functions of x Theorem 2 (Type II) Let h ( y ) , h 1 ( y ) be two continuous functions defined on an interval [ y , y 1 ] , and such that h ( y ) ≤ h 1 ( y ) . Let f ( x, y ) be a continuous function in D = { ( x, y ) ∈ IR 2 : h ( y ) ≤ x ≤ h 1 ( y ) , y ≤ y ≤ y 1 } . Then, the integral of f ( x, y ) in D is given by Z Z D f ( x, y ) dxdy = Z y 1 y " Z h 1 ( y ) h ( y ) f ( x, y ) dx # dy. Slide 6 ’ & \$ % Example type II • Find the R R D f ( x, y ) dxdy for f ( x, y ) = x 2 + y 2 , D = { ( x, y ) ∈ IR 2 : 0 ≤ x ≤ 1 , x 2 ≤ y ≤ x } . Math 20C Multivariable Calculus Lecture 22 4 Slide 7 ’ & \$ % Notice that h ( y ) = y , and h 1 ( y ) = √ y . Then, D = { ( x, y ) ∈ IR 2 : h ( y ) = y ≤ x ≤ h 1 ( y ) = √ y, y ≤ y ≤ y 1 } . Z Z D f ( x, y ) dxdy = Z 1 " Z √ y y ( x 2 + y 2 ) dx # dy, = Z 1 · 1 3 ³ x 3 | √ y y ´ + y 2 ³ x | √ y y ´ ¸ dy, = Z 1 · 1 3 ( y 3 / 2- y 3 ) + y 2 ( y 1 / 2- y ) ¸ dy, = Z 1 · 1 3 y 3 / 2- 1 3 y 3 + y 5 / 2- y 3 ¸ dy, = 1 3 2 5 y 5 / 2 | 1- 1 3 1 4 y 4 | 1 + 2 7 y 7 / 2 | 1- 1 4 y 4 | 1 , = 2 15- 1 12 + 2 7- 1 4 = 9 3 × 5 × 7 . Slide 8 ’ & \$ % • Find the R R D f ( x, y ) dxdy for f ( x, y ) = 1 , D = ½ ( x, y ) ∈ IR 2 : x 2 9 + y 2 4 ≤ 1 ¾ ....
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w8-C - Math 20C Multivariable Calculus Lecture 21 1 Slide 1...

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