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Solution: APPM 1360 Exam 2 Fall 2011
- (a) (10 pts) Find the volume generated by rotating the region in the first quadrant bounded by y = x^2 + 1 and x = 1 about the y-axis using the method of cylindrical shells. (b) (8 pts) Set-up, but do not evaluate, an integral to find the volume of the solid obtained by rotating the region bounded by y^2 = x and x = 2y about the y-axis using the disk/washer method. Simpify the integrand. Solution: (a) (10 pts) Note here r = x and h = x^2 + 1 and so 2πrh∆x = 2π · x · (x^2 + 1)∆x = 2 π(x^3 + x)∆x and so,
V =
0
2 π(x^3 + x) dx = 2π
[
x^4 4
2 2
] 1
0
= 2π
[
+^1
]
= 3 π 2
(b) (8 pts) Note that here R = 2y and r = y^2 , and y^2 = 2y implies intersection points y = 0 and y = 2, and so
V =
∫ (^) b
a
π
[
R^2 − r^2
]
dy =
0
π
[
(2y)^2 − (y^2 )^2
]
dy =
0
π
[
4 y^2 − y^4
]
dy
- (a) (10 pts) Find the length of the part of the curve y^2 = 4x^3 in the first quadrant for 0 ≤ x ≤
(b) (8 pts) Set-up, but do not evaluate, an integral to find the surface area of the surface generated by rotating the curve y = 2x^3 /^2 , 0 ≤ x ≤ 1, rotated about the y-axis. Simplify the integrand. Solution: (a) (10 pts) The part of the curve y^2 = 4x^3 in the first quadrant is y =
4 x^3 = 2x^3 /^2 and y′^ = 3x^1 /^2 and so 1 + (y′)^2 = 1 + 9x and thus
L =
∫ (^) b
a
1 + f ′(x)^2 dx =
0
1 + 9x dx (^) ︸︷︷︸= u=1+9x
1
u du =^1 9
[
u^3 /^2
] 4
1
[8 − 1] = 14
(b) (8 pts) Note that SA =
2 πrdL, where r is the radius and dL is the arclength. Now note that,
in terms of y, r = g(y) and dL =
1 + g′(y)^2 dy, where g(y) =
( (^) y 2
and
g′(y)^2 =
( (^) y 2
=^1
( (^) y 2
and note that 0 ≤ x ≤ 1 implies 0 ≤ y ≤ 2, thus, in terms of y, we have,
SA =
2 πrdL =
0
2 πg(y)
1 + g′(y)^2 dy =
0
2 π
( (^) y 2
( (^) y 2
dy
Now, in terms of x, we have r = x and dL =
1 + f ′(x)^2 dx where f (x) = 2x^3 /^2 and f ′(x)^2 = (3x^1 /^2 )^2 = 9x, thus, in terms of x, we have,
SA =
2 πrdL =
0
2 πx
1 + f ′(x)^2 dx =
0
2 πx
1 + 9x dx =
0
2 πx
1 + 9x dx
- Consider a thin metal plate with density ρ = 3 kg/cm^2 whose shape is bounded by y =^1 2
x^2 and y =
x.
(a) (10 pts) Find the mass of the object. (b) (8 pts ) Set up, but do not evaluate, an integral for the moment about the x-axis, Mx.
Solution: (a) (10 pts) Note that we can find our endpoints by setting the functions equal to each other. 1 2 x
(^2) = √x ⇐⇒ 1 4 x
(^4) = x ⇐⇒ x (^4) = 4x ⇐⇒ x(x (^3) − 4) = 0
giving our endpoints to be 0 and 3
Now, to find the mass we use the formula M = ρA = ρ
∫ (^) b
a
(f (x) − g(x))dx = 3
0
x − x
2 2
dx, and so,
M = 3
∫ √^34
0
x −
x^2 2
dx = 3
[
3 x^
(^32) −
6 x
3
] √^34
0
[
3 (^
6 (^
]
[
3 −^
]
3 = 2 kg
(b) (8 pts) To find the moment about the x-axis we use the formula Mx =
y dm ˜ , where ˜y is given
by ˜y = f^ (x) +^ g(x) 2
=^1
x + x^2 /2) and so,
Mx =
∫ (^) b
a
2 (f^ (x) +^ g(x))^ ·^ ρ(f^ (x)^ −^ g(x))dx^ =^ ρ
∫ (^) b
a
2 (f^ (x)
(^2) − g(x) (^2) )dx
0
x − 1 4
x^4
dx =^3 2
0
x − 1 4
x^4
dx
and so, Mx =^3 2
0
x − 1 4
x^4
dx
- (23 pts) Determine whether the following sequences converge or diverge. For those that converge, find the limit. Justify your answers:
(a)
(n)
lnn n^ }∞ n=
(b) an = (−1)n^ tan
− (^1) (n) en^ (c) (7 pts)^ an^ =^
n! (2n)n^ , n^ ≥^1 Solution: (a) (8 pts) We approximate the sequence with the function f (x) = (x) lnx^ x^. Note that here we have xlim→∞ f^ (x) = “∞^0 ” and so taking the log of both sides yields
ln(y) = ln
(x)
lnx x^ ) = ln( xx )· ln(x) = (ln^ x)
2 x Taking the limit yields,
xlim→∞^ (ln^ x)
2 x
L =′H lim x→∞
2 ln(x) · (^1) x 1
= lim x→∞^ 2 ln^ x x
L =′H lim x→∞
(^2) x^1 1
= lim x→∞^2 x
Therefore ln(y) → 0, and so y = eln(y)^ → e^0 = 1 and so an = (n) lnn^ n^ converges to 1.
and so we can apply the Integral Test: ∫ (^) ∞
10
ke−k^2 dk = lim t→∞
∫ (^) t
10
ke−k^2 dk
and using the u-substitution u = −k^2 ⇒ du = − 2 k dk ⇒ −du/2 = k dk and so,
∫ ke−k^2 dk = − 1 2
eu^ du = −e
u 2
−k^2 2
+ C = −^1
2 ek^2
+ C
and thus,
tlim→∞
∫ (^) t
10
ke−k
2 dk = lim t→∞
2 ek^2
t
10
= lim t→∞
2 et^2
2 e^100 =^
2 e^100
and so the integral converges and thus the series converges by the Integral Test.
Alternately, one could also use a Limit Comparison Test with, for example, the convergent ∑ p-series 1 k^2
, then evaluating the limit and using L’Hˆopital’s Rule succesively yields,
klim→∞
ak bk^ = lim k→∞
k/ek^2 1 /k^2 = lim k→∞
k^3 ek^2
L =′H 0
and so the series converges by Limit Comparison Test.
(c) (7 pts) Using a Limit Comparison with the convergent Geo Series
∑^ ∞
b=
2 −b^ =
∑^ ∞
b=
2 b^
∑^ ∞
b=
)b ,
we have,
blim→∞
ln
1 + 2−b
2 −b
L =′H lim b→∞
− 2 −b^ ln(2)/(1 + 2−b) − 2 −b^ ln(2) = lim b→∞
1 + 2−b^ = 1
now since 0 < 1 < ∞, we have that
∑^ ∞
b=
ln
1 + 2−b
converges by Limit Comparison Test.
