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Solution: APPM 1360 Exam 2 Spring 2012
- (10 pts) Suppose the region enclosed by y = x + ex, the x-axis, the y-axis and x = 1 is rotated about the line x = −1, find the volume of the resulting solid. Solution:
-2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.4 2. -0.
4
(12 pts) Note that we can’t use the disk method since we can’t solve y = x + ex^ for x, so using the shell method, we have r = x + 1 and h = x + ex^ and so
2 πrh∆x = 2π · (x + 1) · (x + ex)∆x = 2π
[
x^2 + x + (x + 1)ex
]
∆x
thus, V =
0
2 π
[
x^2 + x + (x + 1)ex
]
dx
now for
(x + 1)ex^ dx we use integration by parts with u = x + 1 ⇒ du = dx and dv = ex^ ⇒ v = ex which yields (^) ∫ (x + 1)ex^ dx = (x + 1)ex^ −
exdx = (x + 1)ex^ − ex^ = xex so,
V =
0
2 π
[
x^2 + x + (x + 1)ex
]
dx = 2π
[
x^3 3
2 2
] 1
0
= 2π
+^1
= 2π
- (a) (10 pts) Find the length of the curve y^2 = 1 36
(x^2 + 4)^3 for 0 ≤ x ≤ 3. (b) (5 pts) Set up, but DO NOT evaluate an integral to find the surface area of the surface obtained by rotating the curve x = 1 + 2y^2 for 1 ≤ y ≤ 2 about the x-axis. Solution:
(a) (10 pts) The length of a curve y = f (x) for a ≤ x ≤ b is L =
∫ (^) b
a
dy dx
dx. Here we have,
y =^1 6
(x^2 + 4)^3 /^2 ⇒ dy dx
= x 2
x^2 + 4 ⇒ 1 +
dy dx
= 1 + x
2 4
(x^2 + 4) = x
(^4) + 4x (^2) + 4 4
=^1
(x^2 + 2)^2
Then, L =
0
(x^2 + 2)^2 4
dx =^1 2
0
(x^2 + 2) dx =^1 2
[
x^3 3
] 3
0
=^1
[
]
(b) (5 pts) The surface area a curve rotated about the x-axis is SA =
2 πrds where r = y, now if we
partition the y-axis we have SA =
∫ (^) d
c
2 πy
dx dy
dy. Here dx dy
= 4y ⇒ 1 +
dx dy
= 1 + 16y^2 ,
and so the surface area is given by SA =
1
2 πy
1 + 16y^2 dy.
If we partition the x-axis, then SA =
2 πrds where r = f (x), here f (x) =
x − 1 2
for 3 ≤ x ≤ 9 and
we have SA =
∫ (^) b
a
2 πf (x)
dy dx
dx, here dydx =^14
x − 1 ⇒^ 1 +
dy dx
= 1 + (^) 16(x^2 − 1) , and so
the surface area is given by SA =
3
2 π
x − 1 2
1 + (^) 16(x^2 − 1) dx =
3
2 π
x − 1 2 +
16 dx.
- (20 pts) Determine whether the following sequences converge or diverge. For those that converge, find the limit. Justify your answers, show all work:
(a)
n
) 1 /n}∞
n=
(b) an =
n − sin^2 (n) cos^2 (n) + en^ , for^ n^ = 106,^107 ,... Solution:
(a) (10 pts) We approximate the sequence with the function f (n) =
n
) 1 /n
. Note that here we have
nlim→∞ f^ (n) = “0^0 ” which is an indeterminate form, and so taking the log of both sides yields
ln[f (n)] = ln
[(
n
) 1 /n]
ln(1/n) n =^
− ln(n) n
Taking the limit yields,
nlim→∞^ −^ ln(n) n
L =′H lim n→∞
− 1 /n 1
= lim n→∞ − 1 n
Therefore ln[f (n)] → 0, and so f (n) = eln[f^ (n)]^ → e^0 = 1 and so an =
n
) 1 /n converges to 1.
(b) (10 pts) We use the squeeze theorem. Note that
0 ≤ sin^2 (n) ≤ 1 =⇒ 0 ≥ − sin^2 (n) ≥ −1 =⇒
n ≥
n − sin^2 (n) ≥
n − (^1) ︸︷︷︸≥ since n > 1
so 0 ≤
n − sin^2 (n) ≤
n, and cos^2 (n) ≥ 0 implies en^ + cos^2 (n) ≥ en, so, (^) en (^) + cos^12 (n) ≤ (^) e^1 n , putting all this together yields
n − sin^2 (n) cos^2 (n) + en^ ≤
n cos^2 (n) + en^ ≤
n en^ for^ n^ = 106,^107 ,...
Furthermore, lim n→∞
n en
L =′H lim n→∞
nen^
= 0, thus the sequence converges to 0 by the Squeeze Theorem.
- Consider a thin metal plate with density ρ = 2 lb/ft^2 whose shape is the region enclosed by the curve x =
y + 1 , the^ x-axis, the^ y-axis, and the line^ y^ = 2. (a) (10 pts ) Find the total mass of the plate by partitioning the y-axis. (b) (5 pts) Set up, but DO NOT evaluate, an integral in terms of y to find the moment of the plate about the x-axis. Solution: (a) (10 pts) The graph:
-0.5 -0.25 0 0.25 0.5 0.75 1 1.25 1. -0.
2
Note that we partition the y-axis to avoid needing two separate integrals.
The total mass is M =
∫ (^) b
a
ρ dA =
0
y + 1 dy^ = 2 [ln^ |y^ + 1|]
2 0 = 2 ln 3 = ln 9.
(b) (5 pts) Here ˜y = y, so the moment about the x-axis is Mx =
∫ (^) b
a
yρ dA ˜ =
0
2 y y + 1
dy.
- (20 pts) Answer either Always True or False. Do NOT justify your answer.
(a) If 0 ≤ an+10 ≤ bn for n = 1, 2 ,... and
∑^ ∞
n=
bn converges, then
∑^ ∞
n=
an converges.
(b) If
∑^ ∞
n=
an converges and if an 6 = 1 and an > 0 for all n then
∑^ ∞
n=
an 1 − an^ converges.
(c) The function y = − ln[ecos(x)] is a solution to the differential equation csc(x) dy dx
= ey+cos(x).
(d) Let f and b be constants with 0 < f < b, then the sequence
(f m^ + bm)^1 /m
m=1 converges to^ b. Solution: (a) A.T. (b) A.T. (c) A.T. (d) A.T.
Comments: (a) This implies 0 ≤
∑^ ∞
n=
an ≤
∑^ ∞
n=
bn and so
∑^ ∞
n=
an converges by the Direct Comparison Test, thus
∑^ ∞
n=
an =
∑^10
n=
an +
∑^ ∞
n=
an converges.
(b) First note that since
∑^ ∞
n=
an is convergent we have (^) nlim→∞ an = 0, so there exists some number N for
which 0 < an < 1 for all n > N , (so an 1 − an
> 0 for all n > N ), now using the Limit Comparison Test we
have (^) nlim→∞
an/(1 − an) an^ =^ nlim→∞
1 − an^ = 1, so
∑^ ∞
n=N
an 1 − an^ converges by the Limit Comparison Test and thus ∑^ ∞
n=
an 1 − an
converges.
(c) Note that
dy dx =
sin(x)ecos(x) ecos(x)^
= sin(x) and ey^ = e−^ ln[e
cos(x)] = eln[(e
cos(x))− (^1) ]
ecos(x)^
, so
csc(x)
dy dx = csc^ x^ sin(x) = 1 and^ e
y+cos(x) (^) = eyecos(x) (^) = 1 ecos(x)^ e
cos(x) (^) = 1 and so, csc(x) dy dx =^ e
y+cos(x)
thus y = − ln[ecos(x)] is a solution of csc(x) dy dx
= ey+cos(x)
(d) Note (^) mlim→∞ (f m^ + bm)^1 /m^ = (^) mlim→∞ b
f b
)m
) 1 /m = b since f < b implies
f b <^ 1.
