# Chapt_09_2 - 9 Vector Calculus DO NOT USE THIS PAGE 504...

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9.13 Surface Integrals EXERCISES 9.13 Surface Integrals 1. Letting z = 0, we have 2 x +3 y = 12. Using f ( x, y )= z =3 1 2 x 3 4 y we have f x = 1 2 , f y = 3 4 ,1+ f 2 x + f 2 y = 29 16 . Then A = Z 6 0 Z 4 2 x/ 3 0 p 29 / 16 dy dx = 29 4 Z 6 0 µ 4 2 3 x dx = 29 4 µ 4 x 1 3 x 2 ¯ ¯ ¯ ¯ 6 0 = 29 4 (24 12) = 3 29 . 2. We see from the graph in Problem 1 that the plane is entirely above the region bounded by r = sin2 θ in the Frst octant. Using f ( x, y z 1 2 x 3 4 y we have f x = 1 2 , f y = 3 4 , 1+ f 2 x + f 2 y = 29 16 . Then A = Z π/ 2 0 Z sin2 θ 0 p 29 / 16 rdrdθ = 29 4 Z 2 0 1 2 r 2 ¯ ¯ ¯ ¯ θ 0 = 29 8 Z 2 0 sin 2 2 θdθ = 29 8 µ 1 2 θ 1 8 sin4 θ ¯ ¯ ¯ ¯ 2 0 = 29 π 32 . 3. Using f ( x, y z = 16 x 2 we see that for 0 x 2 and 0 y 5, z> 0. Thus, the surface is entirely above the region. Now f x = x 16 x 2 , f y =0, f 2 x + f 2 y =1+ x 2 16 x 2 = 16 16 x 2 and A = Z 5 0 Z 2 0 4 16 x 2 dx dy =4 Z 5 0 sin 1 x 4 ¯ ¯ ¯ ¯ 2 0 dy Z 5 0 π 6 dy = 10 π 3 . 4. The region in the xy -plane beneath the surface is bounded by the graph of x 2 + y 2 =2. Using f ( x, y z = x 2 + y 2 we have f x =2 x , f y y f 2 x + f 2 y =1+4( x 2 + y 2 ). Then, A = Z 2 π 0 Z 2 0 p 1+4 r 2 = Z 2 π 0 1 12 (1+4 r 2 ) 3 / 2 ¯ ¯ ¯ ¯ 2 0 = 1 12 Z 2 π 0 (27 1) = 13 π 3 . 5. Letting z = 0 we have x 2 + y 2 = 4. Using f ( x, y z ( x 2 + y 2 )wehave f x = 2 x , f y = 2 y f 2 x + f 2 y x 2 + y 2 ). Then A = Z 2 π 0 Z 2 0 p r 2 = Z 2 π 0 1 3 r 2 ) 3 / 2 ¯ ¯ ¯ ¯ 2 0 = 1 12 Z 2 π 0 (17 3 / 2 1) = π 6 (17 3 / 2 1) . 505
9.13 Surface Integrals 6. The surfaces x 2 + y 2 + z 2 = 2 and z 2 = x 2 + y 2 intersect on the cylinder 2 x 2 +2 y 2 =2 or x 2 + y 2 = 1. There are portions of the sphere within the cone both above and below the xy -plane. Using f ( x, y )= p 2 x 2 y 2 we have f x = x p 2 x 2 y 2 , f y = y p 2 x 2 y 2 ,1+ f 2 x + f 2 y = 2 2 x 2 y 2 . Then A " Z 2 π 0 Z 1 0 2 2 r 2 rdrdθ # 2 Z 2 π 0 p 2 r 2 ¯ ¯ ¯ 1 0 2 Z 2 π 0 ( 2 1) =4 π 2( 2 1) . 7. Using f ( x, y z = p 25 x 2 y 2 we have f x = x p 25 x 2 y 2 , f y = y p 25 x 2 y 2 f 2 x + f 2 y = 25 25 x 2 y 2 . Then A = Z 5 0 Z 25 y 2 / 2 0 5 p 25 x 2 y 2 dx dy =5 Z 5 0 sin 1 x p 25 y 2 ¯ ¯ ¯ ¯ 25 y 2 / 2 0 dy Z 5 0 π 6 dy = 25 π 6 . 8. In the Frst octant, the graph of z = x 2 y 2 intersects the xy -plane in the line y = x . The surface is in the Frt octant for x>y . Using f ( x, y z = x 2 y 2 we have f x x , f y = 2 y f 2 x + f 2 y =1+4 x 2 +4 y 2 . Then A = Z π/ 4 0 Z 2 0 p 1+4 r 2 = Z 4 0 1 12 (1+4 r 2 ) 3 / 2 ¯ ¯ ¯ ¯ 2 0 = 1 12 Z 4 0 (17 3 / 2 1) = π 48 (17 3 / 2 1) . 9. There are portions of the sphere within the cylinder both above and below the xy -plane. Using f ( x, y z = p a 2 x 2 y 2 we have f x = x p 1 2 x 2 y 2 , f y = y p a 2 x 2 y 2 , 1+ f 2 x + f 2 y = a 2 a 2 x 2 y 2 . Then, using symmetry, A " 2 Z 2 0 Z a sin θ 0 a a 2 r 2 # a Z 2 0 p a 2 r 2 ¯ ¯ ¯ a sin θ 0 a Z 2 0 ( a a p 1 sin 2 θ ) a 2 Z 2 0 (1 cos θ ) a 2 ( θ sin θ ) ¯ ¯ ¯ 2 0 a 2 ³ π 2 1 ´

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