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Solution: APPM 1360 Exam 2 Spring 2011
- (a) (3 pts) Graph the region bounded by y = x − 2 and x = y^2.
(b) (12 pts) Find the volume of the solid created by revolving the region bounded by y = x − 2 and x = y^2 about the line y = 3. Solution: (a) The graph
-0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4 4.4 4.
-2.
-1.
-0.
enclosed region
(b) Here ∆V = 2πrh with r = 3 − y and h = (y + 2) − y^2 and so
V =
− 1
2 π (3 − y)
y + 2 − y^2
dy = 2π
− 1
y^3 − 4 y^2 + y + 6
dy = 2π
[
y^4 4
− 4 y
3 3
2 2
2
− 1
]
=^45 π 2
- (a) (2 pts) Graph the region bounded by the curve x^3 = 4/y, and the lines x = 1 and y = 1/2. (b) (8 pts) Set up, but do not solve, an integral (or integrals) to find the volume of the solid generated by revolving the region bounded by the curve x^3 = 4/y, and the lines x = 1 and y = 1/2 about the line x = −e, using the disk/washer method Solution: (a) The graph
-0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4 -0.
4
enclosed region
(b) Here ∆V = π[R^2 − r^2 ] with R = 3
4 /y + e and r = 1 + e and so
V =
1 / 2
π
[(
√ (^34) /y + e
− (1 + e)^2
]
dy
- (10 pts) Solve the following differential equations: (a) 2
xy
dy dx = 1^ (b)^
dx dt =^ e
t−x
Solution:
(a) Separating yields y^1 /^2 dy = x
− 1 / 2 2
dx and so integrating and rearranging terms yields
y^3 /^2 =^3 2
x^1 /^2 + C, and so y =
x^1 /^2 + C
(b) Separating yields exdx = etdy and so integrating and rearranging terms yields
ex^ = et^ + C, and so x = ln(et^ + C)
- (20 pts) Consider a thin metal plate with density ρ(x) = 2 covering the region bounded by the curve y = arctan(x), 0 ≤ x ≤ 1 and the x-axis.
(a) Find the mass of the thin metal plate. (b) Set-up, but do not solve, an integral to find the moment about the x-axis, Mx, of the thin metal plate.
Solution: (a) Note that here M =
ρ(x)dA =
0
2 arctan(x)dx, where this integral is done using integration by parts with u = arctan(x) and dv = dx, then du = dx/1 + x^2 and v = x, so
M =
ρ(x)dA =
0
2 arctan(x)dx (^) ︸︷︷︸= using I.B.P
2 x arctan(x)
1
0
0
2 x 1 + x^2 dx
Now, we use a u-substitution to solve the last integral with u = 1 + x^2 , then du = 2xdx and ∫ 2 x 1 + x^2
dx (^) ︸︷︷︸= u=1+x^2
du u
= ln |u| = ln(1 + x^2 )
so,
M =
0
2 arctan(x)dx = 2x arctan(x)
1
0
− ln(1 + x^2 )
1
0
= 2 arctan(1) − ln(2) =
2 π 4 −^ ln(2)
(b) Here, if we partition the x-axis, then the center of each strip is (˜x, y˜) = (x, arctan(x)/2), thus the moment about the x-axis is given by
Mx =
˜ydm =
y ˜ · ρ(x)dA =
0
arctan(x) 2 ·^ 2 arctan(x)dx^ =
0
arctan^2 (x)dx
(g) Here we have a geometric with,
∑^ ∞
n=
5(−1)n 32 n^ =
∑^ ∞
n=
)n
so a = 5/81 and r = − 1 /9 and |r| < 1 and so
∑^ ∞
n=
5(−1)n 32 n^ =^
90 =^
thus this is a convergent series which converges to 1/18.
