16 MULTIPLE INTEGRALS

(b)
3. (a)
CHAPTER 16 SAMPLE EXAM SOLUTIONS
∫ π/2 ∫ π/2 ∫
( 1/ 1+sin 2 y ) ∫ π/2 ∫ π/2
(
)
0 0 1/ ( 1+sin 2 x ) dzdx dy = 1
0 0 1 + sin 2 y − 1
1 + sin 2 dx dy
x
∫ π/2 ∫ π/2
( ) ∫
1
π/2 ∫ π/2
( )
1
=
0 0 1 + sin 2 dx dy −
y
0 0 1 + sin 2 dx dy
x
= π2
4 √ 3 − π2
4 √ 3 = 0
(b)
6
Type I
Type II
7
4. ∫ 2π
0
∫ 1
0 r2 dr dθ = ∫ 1
−1
∫ √ 1−x
√1−x 2 √
x 2 + y 2 dydx
2
−
5. ∫ 2π ∫ 2
0 1 cos ( r 2) rdrdθ = π (sin 4 − sin 1)
6. x 2 + 2y 2 = 1
(a) r 2 ( cos 2 θ + 2sin 2 θ ) = r 2 ( 1 + sin 2 θ ) = 1, r ≥ 0
(b)
∫ 2π
∫ 1
/√1+sin 2 θ
0
0
rdrdθ
7. Since the parallelepiped has volume 60, we have ∫∫∫ R
10 dV = 600.
∫∫∫R xdV = 12 ∫ ( )
5
0 xdx= 12 252
= 150
8. (a) ∫ 1 ∫ 1 ∫ xy
0 −1 0
1 dz dx dy = ∫ 1 ∫ 1
0 −1 xydx dy = ∫ [ ]
1 1 12
0
x 2 y dy = 0. The region between z = 0and
−1
z = xy in the first quadrant is above the xy-plane, while a symmetric region is below the xy-plane in
the second quadrant.
937