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20Chw6

20Chw6 - MATH 20C graded homework 6(20 points 15.1 26(6...

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MATH 20C graded homework 6. (20 points) 15.1 # 26 (6 points): Evaluate the iterated integral R π/ 4 0 R π/ 2 0 cos (2 x + y ) dydx . Solution: Z π/ 4 0 Z π/ 2 0 cos(2 x + y ) dydx = Z π/ 4 0 sin(2 x + y ) π/ 2 0 dx = Z π/ 4 0 sin(2 x + π 2 ) - sin(2 x ) dx = Z π/ 4 0 sin(2 x ) cos( π 2 ) + cos(2 x ) sin( π 2 ) - sin(2 x ) dx = Z π/ 4 0 cos(2 x ) - sin(2 x ) dx = 1 2 (sin(2 x ) + cos(2 x )) π/ 4 0 = 1 2 sin( π 2 ) + cos( π 2 ) - sin(0) - cos(0) = 0 . 15.2 # 12 (6 points): Find the double integral of f ( x, y ) = x 3 y over the region between the curves y = x 2 and y = x (1 - x ) . Solution: First find the points of intersection of the two curves by solving x 2 = x (1 - x ), giving x = 0 , 1 2 . . . . . 1 1 (0 , 0) ( 1 2 , 1 4 ) ( Remark: Making a rough sketch of the region of integration will help you figure out whether x 2 or x (1 - x ) should be the bottom limit.) So we want to calculate Z 1 2 x =0 Z x - x 2 y = x 2 x 3 y dydx = Z 1 2 x =0 x 3 1 2 y 2 x - x 2 y = x 2 ! dx = Z 1 2 x =0 1 2 x 3 ( ( x - x 2 ) 2 - x 4 ) dx = 1 2 Z 1 2 x =0 x 5 - 2 x 6 dx = 1 2 1 6 x 6 - 2 7 x 7 1 2 x =0 = 1 2 1 6 1 64 - 2 7 1 128 = 1 5376 .

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15.2 # 52 (8 points): Evaluate R R D x dA for D in Figure 24. Solution: There are multiple ways to break down this question. The region of integration D is half an annulus (lying in the half plane where
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