Mu Multip ltiple le Int Integ egrals rals Double Integrals: - PowerPoint PPT Presentation

Mu Multip ltiple le Int Integ egrals rals

Double Integrals:     y d b d x b       f x y dydx ( , ) f x y dy dx ( , )       a c x a y c Important properties of the double integral:        , , a f x y dA a f x y dA R                 f x y , g x y , dA f x y dA , g x y dA ,   R R R            f x y dx dy , f x y dx dy , f x y dx dy , R R R 1 2

Example 2 3  2 Evaluate the iterated integral: x ydxdy 1 0    y 2 2 3 3    Solution:  2 2   x ydxdy x ydx dy    1 0 y 1 0 3   2 2 3 x     y   dy 9 ydy   3 1 1 0 2   2 y 27     9   2 2 1

Evaluating Double Integrals over General Regions         h y d g x 2 b 2                 f ( , x y ) dA f x y dx , dy f ( , x y dA ) f x y , dy dx             D c h y   D a g x 1 1

Example   ( 2 ) D x y dA Evaluate  2   2 D y : 2 x and y 1 x . Solution:  2 1 1 x       ( x 2 ) y dA ( x 2 y dy ) dx D   2 x 1 y 2 x  2 1 x 1      2 xy y dx    2 1 2 x 1 32        2 3 4 (1 x 2 x x 3 x ) dx 15  1

Example 2 1    2 y e dy dx Evaluate 0 x /2 Solution:     D y : x /2, y 1, x 2, x 0. y y  1 2 y 2 1 1  2  2     y  y e dy dx e dx dy  y x / 2 0 / 2 0 0 x x  1 x 2 y   x   2 2   y   x e dy    x 0 0 1 2 1      2 1     y        y 1     2 y e dy e e 1 1     e 0 0

The Double Integral in Polar Coordinate:     x r cos y r sin  rdrd    2 2 dA x y r

Example   2 (3 x 4 y ) dA in the upper half-plane Evaluate D 2  2  2  2  bounded by the circles D x : y 1 and x y 4. Solution:  2          2 2 2 (3 x 4 y ) dA (3 cos r 4 r sin ) rdrd R    0 r 1  2        2 3 2 (3 r cos 4 r sin ) drd    0 r 1   2      3 4 2 ( cos sin ) r r d 1 0    15 15               2 (7cos 15sin ) d (7cos (1 cos2 )) d 2 2 0 0

Applications of Double Integrals: (1) Calculating the area of a plane region:   A dA D (2) Calculating the Volumes:   V f x y dA ( , ) . D

Example Calculate the area of a region bounded the curves:   2  y 2 x , y x . Solution:  2 1 2 x 1  2        2 x   A dA dy dx y dx x   2 x 2 D 1   1   3 2 x x        2   2 x x dx 2 x   3 2     2 2 1 1 8 4 27        2 4 3 2 3 2 6

Example Calculate the volume of a solid bounded by the surfaces:       x 0, y 0, x y z 1, z 0. Solution:         V f x y dA ( , ) 1 x y dA D D  1 1 x        1 x y dydx 0 0  1 x     2 1 1  2 1 x y                 1 x y dx 1 x 1 x dx   2 2   0 0 0 1       2 3 1   1 x 1 x 1       dx    2 6 6   0 0

Example Find the volume of the solid bounded by the paraboloid:   2  2  1 , 0. z x y z Solution:       2 2 V f x y dA ( , ) (1 x y ) dA D D  2 1  2 1        2    (1 ) 3 r rdrd d ( r r dr ) 0 0 0 0 1     2 4 r r    2     2 4 2 0