1 Calculus III Exam 2 Practice Problems - Solutions

Calculus III, Exam 2: Review Solutions 4
(c) ∫ 1
0
1
(d) ∫ 1
0
∫ 1
y sin(x 2 )dxdy = ∫ 1 ∫ √ x
y sin(x 2 )dydx = ∫ 1
y 2 0 0
[1 − cos 1]
2 [− 1 2 cos(x2 )] 1 0 = 1 4 [− cos(1) + cos 0] = 1 4
∫ x
2 ∫ y
0 0 y2 zdzdydx = ∫ 1 ∫ x
2
y 2 [ 1 0 0 2 z2 ] y 0 dydx = ∫ 1
0
[ 1
0 sin(x2 )[ 1 √ x
2 y2 ]
∫ x
2
0
0 dx = ∫ 1
0 sin(x2 )[ 1 2 x]dx =
1
2 y4 dydx = ∫ 1
0 [ 1
10 y5 ] x2
0 dx = ∫ 1 1
0 10 x1 0dx =
110 x11 ] 1 0 = 1
110
(e) ∫ 1 ∫ 1
0 x ex/y dydx = ∫ 1 ∫ y
0 0 ex/y dxdy = ∫ 1
0 [yex/y ] y 0 dy = ∫ 1
0 [ye − y]dy = [ e 2 y2 − 1 2 y2 ] 1 0 = 1 2
(e − 1)
(f) ∫ 1 ∫ 1√y ∫ y
0 0 xydzdxdy = ∫ 1 ∫ 1√y
xy[z] y 0
0 dxdy = ∫ 1 ∫ 1√y
xy 2 dxdy = ∫ 1
0
0 y2 [ 1 2 x2 ] 1 √ y
dy = ∫ 1
0 [ 1 2 y2 −
1
2 y3 ]dy = [ 1 6 y3 − 1 8 y4 ] 1 0 = 1 6 − 1 8 = 1
24
16. Use differentials to estimate the amount of tin in a closed tin can with diameter 8cm and height
12 cm if the tin is 0.04 cm thick.
Solution: The volume of tin is the volume of the exterior of the can minus the volume of the
interior of the can. The volume of a cylinder of radius r and height h is V = πr 2 h, so ∆V ≈ dV =
V r dr + V h dh = [2πrh][dr] + [πr 2 ][dh] = [2π(4)(12)][0.04] + [π(4) 2 ][0.04] = 112π
25 .
17. Find the volume under the cone z = √ x 2 + y 2 and above the ring 4 ≤ x 2 + y 2 ≤ 25.
∫ 5 ∫ r
r=2 z=0 rdzdrdθ = ∫ 2π ∫ 5
0 2 r[z]r 0drdθ = ∫ 2π
0
Solution: V = ∫ 2π
θ=0
( 125
3 − 8 3
)[2π] = 78π.
∫ 5
2 r2 drdθ = ∫ 2π
0 [ 1 3 r3 ] 5 2dθ =
18. Find the average height of the paraboloid z = x 2 + y 2 over the rectangle R = {(x, y) : 0 ≤ x ≤
2, 0 ≤ y ≤ 1}.
∫∫
1
Solution: The average height is
Area(R) R zdA = ∫ 1 2 ∫ 1 ∫
2 0 0 (x2 + y 2 )dydx = 1 2
2 0 [x2 y + 1 3 y3 ] 1 0dx =
∫ 2
0 [x2 + 1 3 ]dx = 1 2 [ 1 3 x3 + 1 3 x]2 0 = 1 6 [(8 + 2) − (0 + 0)] = 5 3 .
1
2