# Test 2 – Math 2300 – Fall 2012

```Test 2 – Math 2300 – Fall 2012
On my honor as a University of Colorado at Boulder student I have neither given nor
received unauthorized assistance on this exam.
Name:
001 Scherer . . . . . . . . . (9am)
002 Davison . . . . . . . . (10am)
003 Gorokhovsky . . (11am)
004 Gorokhovsky . . (12pm)
005 Moore . . . . . . . . . . (2pm)
006 Rosenbaum . . . . . . (3pm)
In order to receive full credit your answer must be complete, legible and correct. Show all of
your work, and give clear explanations.
Problem Max. points Points
1
15
2
10
3
20
4
10
5
10
6
20
7
15
Total
100
1
2
1. (15 pt) Find the area of the region in the polar coordinates bounded by the curve r =
and rays θ = 0 and θ = π/2.
Z
β
α
Z π/2 √
Z π/2
2
1 2
1
1
r dθ =
2θ + 1 dθ =
(2θ + 1)dθ
2
2
2
0
0
π/2
Z π/2
1
θ2 θ (π/2)2 π/2
=
θ + dθ =
+ =
+
2
2
2
2
2
0
0
√
2θ + 1
3
2. (10 pt) Find the equation (in terms of x and y) of the tangent line to the curve r = θ at θ = π/2.
x = r cos(θ) = θ cos(θ) and y = r sin(θ) = θ sin(θ)
So
dy
dx
= cos(θ) − θ sin(θ) and
= sin(θ) + θ cos(θ)
dθ
dθ
When θ = π/2,
dx
π
dy
= − and
=1
dθ
2
dθ
and (x, y) = (0, π/2). The tangent line is therefore
1
−2
y=
x + π/2 =
x + π/2
−π/2
π
4
3. A metal plate, with constant density 1 g/cm2 , has a shape bounded by the curve y = x2 and the
line y = 1; and x, y are in cm.
(a) (10 pt) Find the total mass of the plate (include units).
The area is equal to the area of a rectangle with dimensions 1 &times; 2 minus the area under the
curve y = x2 from x = −1 to x = 1, i.e.,
 
1
Z 1
3
1
x
−1
2
4
2
−
x dx = 2 −   = 2 −
=2− =
2−
3
3
3
3
3
−1
−1
Therefore, since the density is 1 g/cm2 the mass is
4
3
g.
(b) (10 pt) Find the coordinates of the center of mass of the plate.
Since the plate has constant density and is symmetric about the y-axis, x = 0.
√
A slice perpendicular to the y-axis has length 2 y and such slices can be taken from 0 to
1, so
R1
R1
Z
√
δA
ydy
1 &middot; 2( y)ydy
3 1 3/2
y
0
0
y=
=
=
2y dy
Total Mass
4/3
4 0
1
12
3 2 5/2 = 2 y =
4 5
20
0
5
4. (10 pt) Find the length of the curve given by parametric equations
x(t) = 2t, y(t) =
t2
− ln t, 1 ≤ t ≤ 2.
2
x0 (t) = 2 and y 0 (t) = t −
1
t
So
2
1
1
1
2
(x (t)) + (y (t)) = 4 + t − 2 + 2 = t + 2 + 2 = t +
t
t
t
and therefore for 1 ≤ t ≤ 2,
s
2
p
1
1
0
2
0
2
=t+
(x (t)) + (y (t)) =
t+
t
t
0
2
0
2
2
Thus, the arclength is
Z 2p
Z
(x0 (t))2 + (y 0 (t))2 dt =
1
1
2
2
1
1
t2
3
t + dt = + ln |t| = 2 + ln |2| − = + ln |2|
t
2
2
2
1
6
5. (10 pt) Determine if the following improper integral
Z∞
√
x
dx
x3 − 1
2
When x ≈ ∞, 1 is insignificant compared to x3 so
x
x
1
x
√
≈ √ = 3/2 = 1/2
x
x
x3 − 1
x3
R ∞ dx
So we predict this integral will diverge because 1 x1/2 diverges. Therefore, we wish to find some
function smaller than √xx3 −1 whose integral diverges on [2, ∞).
But
x
x
1
√
≥ √ = 1/2
3
x3 R x
R ∞ dx
R ∞ dxx − 1
∞
Since 1 x1/2 diverges so does 2 x1/2 and therefore 2 xdx
1/2 diverges.
7
6. For each of the series below determine if it is absolutely convergent, conditionally convergent or
(a) (10 pt)
∞
X
1
√
2
n +1
n=1
Note that
√
1/n
n2 + 1 p
√
= 1 + 1/n2
=
n
1/ n2 + 1
p
P
1
So limn→∞ 1 + 1/n = 1. Therefore, since ∞
n=1 n diverges so does the original sum.
(b) (10 pt)
∞
X
(−1)n
n ln(n)
n=2
1
1
The sequence n ln(n)
is decreasing and limn→∞ n ln(n)
= 0. So by the alternating series test,
the series converges.
On the other hand by using substitution with w = ln(x) we get
Z ∞
dx
= lim ln | ln(b)| − ln | ln(2)| = ∞
x ln(x) b→∞
2
P
1
so that ∞
n=2 n ln(n) diverges. So the series is conditionally convergent.
8
7. Consider the region bounded by the graph of y = ex , line y = e and the y-axis.
(a) (10 pt) Find the volume of the solid obtained by rotating this region about the x-axis
Cross sections are washers with outer radius e and inner radius ex and thickness dx. Thus,
cross sections have area (πe2 − π (ex )2 )dx = π(e2 − e2x )dx. Since these cross sections are
realized for 0 ≤ x ≤ 1, the volume is
1
Z 1
1 2 1
1 2 1
1
2
2x
2x 2
2
π e − e dx = π e x − e
e −
=π
e +
=π e −
2
2
2
2
2
0
0
(b) (5 pt) Find the volume of the solid obtained by rotating this region about the y-axis
A cross section is a thin cylinder with radius ln(y) and height dy. These cross sections have
volume π(ln(y))2 dy and are taken from y = 1 to y = e. Thus, the volume is
Z e
π(ln(y))2 dy
1
We can evaluate this by the substitution u = ln(y) so that y = eu and dy = eu du. So
Z
Z
2
(ln(y)) dy = u2 eu du
Z
By using integration by parts we get
Z
Z
2 u
2 u
u
2 u
u
u e = u e − 2ue du = u e − (2ue − 2eu ) = u2 eu − 2ueu + 2eu = (ln(y))2 y − 2 ln(y)y + 2y
So
Z
1
e
e
2
2
π(ln(y)) dy = π((ln(y)) y − 2 ln(y)y + 2y)
= π(e − 2e + 2e − 2) = π(e − 2)
Other solutions are possible.
1
```