Download Exam 1 Fa 2013

Problem 1
[18 pts]
True or False. Give a brief explanation or example to justify your answer.
a) [3 pts] Given a solid generated by revolving a region about the line x = 3, if we are
using the shell method to compute its volume, then we would integrate with respect to x.
True: The line x = 3 is parallel to the y axis. When you use shells the variable of
integration is opposite to the axis of rotation, so we should integrate over x if we want to
use shells
b) [3 pts] Given a spring that obeys Hooke’s Law, the work required to stretch the
spring from equilibrium to 1 cm is the same as the work required to stretch the spring
from 1 cm to 2 cm.
False: Given a fixed spring constant k, the spring from 1 cm to 2 cm is given by
work required to stretch the spring from
Z 0.02
k((0.02)2 − (0.01)2
3k
equilibrium to 1 cm is given by
kx
=
=
.
Z 0.01
2
20, 000
0.01
k(0.01)2
k
=
.
kx =
2
20, 000
0
Thus the work streaching the spring from 1
Meanwhile, the work required to stretch the cm to 2 cm is greater.
Z
Z
6
3
c) [3 pts]
sin θ cos θ dθ = (u6 − u8 ) du where u = sin θ.
True:
Z
6
3
Z
sin6 θ(1 − sin2 θ) cos θ dθ
R
Let u = sin θ. Then du = cos θ dθ. So our integral is equal to (u6 − u8 ) du, as desired.
Z
1
d) [3 pts]
dx = ln | cos x| + C
cos x
sin θ cos θ dθ =
False:
Z
1
dx =
cos x
Z
sec x dx = ln |sec x + tan x| + C 6= ln |cos x| + C
Alternatively one could differentiate ln |cos x| + C to show the statement was false.
Z 1
e) [3 pts]
x3 dx represents the area of the region bounded by the graph of y = x3
−1
and the x-axis over the interval [−1, 1].
False:
Z
1
−1
Z
f ) [3 pts]
0
π/4
x3 dx =
x4 1
1 1
= − = 0.
4 −1 4 4
π
sec x dx is the length of the curve y = ln(sec x) on the interval [0, ].
4
True: Arc length of the curve y(x) on the Thus arc length is given by
interval [0, π/4] is given by
Z πp
Z π√
4
4
2
1 + tan x dx =
sec2 x dx
Z πp
4
0
0
1 + (y 0 (x))2 dx.
Z π/4
0
=
sec x dx.
1
0
0
y (x) =
sec x tan x = tan x.
sec x
Problem 2
[22 pts] Areas and Volumes. Give the exact values. You do not need to simplify your
final answer.
a) [10 pts] Find the area of the region bounded by the graphs x = y+1 and x = y 2 −y−7.
Solution: Let us equate the two equations the curve y 2 − y − 7. Thus
to identify their intersection pts.
Z 4
y + 1 − (y 2 − y − 7) dy
Area =
−2
Z 4
−y 2 + 2y + 8 dy
=
y + 1 = y2 − y − 7
−2
4
2
−y 3
y2
0 = y − 2y − 8
=
+ 2 + 8y 3
2
−2
0 = (y + 2)(y − 4)
−43
42
=
+2 +8·4
3
2
−(−2)3
(−2)2
−
+2
+ 8 · (−2)
3
2
100
Thus our region is contained in the interval
=
3
[−2, 4]. By evaluating each curve at the pt
y = 0, we see that the curve x = y + 1 has
larger x values on the interval [−2, 4] than
b) [12 pts] Consider the region R bounded by the graph of y = sin (3x) and the x-axis
π
over the interval [0, ]. Find the volume of the solid whose base is the region R and whose
3
cross sections perpendicular to the x-axis are squares.
Solution: On the interval [0, π3 ] sin(3x) ≥ over the bounds of our object. Thus
0. Thus for a fixed x the side length of the
Z π
3
square cross section perpendicular to the x V olume =
sin2 3x dx
axis through that point is given by sin(3x)
0
Z π
and our formula for cross sectional area at
3 1
=
(1 − cos 6x) dx
the point x is given by
0 2
π
1
sin 6x 3
=
x−
2
6
0
π
2
sin
6
1
π
sin 6 · 0
3
CSA(x) = sin 3x.
=
−
− 0−
2 3
6
6
π
=
6
By the generalized slicing method volume
is given by integrating cross sectional area
Problem 3
[20 pts] A tank is shaped like a parabolic bowl. It is formed by revolving the graph of
y = 4x2 for 0 ≤ x ≤ 3 (in meters) about the y-axis. The tank is filled with water to a
height of 30 meters. How much work is required to pump all of the water to an exit pipe
at the top of the tank? [Note: The density of water is 1000 kg/m3 .]
Solution: Work for pumping a liquid out of a tank is given by the equation
Z b
W ork =
ρ · g · CSA(y) · D(y) dy.
a
Here [a, b] is the interval where the liquid exists. ρ is the density of the liquid. g is the
gravitational constant. CSA(y) is the area of the horizontal cross section at the height y.
D(y) is the distance the cross section at height y has to travel.
In our case the tank is filled up to 30m so our bounds of integration are a = 0 and b = 30.
ρ = 1000 and g ≈ 9.8. The tank is formed as a rotational solid with axis of rotation the y
axis. Thus the cross sections are circles and by solving the relation y = 4x2 for y we find
the radius as a function of height:
r
y
.
r(y) =
4
Thus
y
CSA(y) = π .
4
a
The height of the tank is 36 meters . Thus the distance the yth cross section has to travel
to pump out of the tank is 36 − y. Thus
Z 30
πy
1000 · 9.8 ·
W ork ≈
(36 − y) dy
4
0
Z
9800π 30
36y − y 2 dy
=
4
0
30
9800π
y3
2
=
18y −
4
3 0
9800π
303
2
=
18(30) −
4
3
≈ 69272118 J
a
Plug in x = 3 into the equation y = 4x2
Problem 4
[20 pts]
Evaluate the following integrals. Show your work.
Z
a) [10 pts]
x8 ln x dx
Solution: We will solve the integral by integrating by parts. Let u = ln(x) and dv =
9
x8 dx. Then du = x1 dx and v = x9 , and our integral becomes:
Z
3
Z 9
x 1
x9
ln(x) −
· dx
x ln(x) dx =
9
9 x
Z
x9
1
=
ln(x) −
x8 dx
9
9
9
1 x9
x
ln(x) − ·
+C
=
9
9 9
8
w2 − 3
b) [10 pts]
dw
0 w+1
Solution: Let u = w + 1. Then w = u − 1 and dw = du. At w = 0 u = 1 and at w = 3
u = 4. Thus
Z 3 2
Z 4
w −3
(u − 1)2 − 3
dw =
du
u
0 w+1
1
Z 4 2
u − 2u + 1 − 3
=
du
u
1
Z 4 2
u − 2u − 2
=
du
u
1
Z 4 2
u
2u 2
=
−
− du
u
u
1 u
Z 4
2
=
u − 2 − du
u
1
4
2
u
− 2u − 2 ln |u|
=
2
1
2
2
4
1
=
− 2 · 4 − 2 ln |4| −
− 2 · 1 − 2 ln |u|
2
2
Z
Problem 5
[20 pts]
Consider the region under the graph y =
where a > 0 is a constant.
1
over the interval 0 ≤ x ≤ 1,
+1
ax2
a) [12 pts] Set up an integral (or integrals) that represents the volume of the solid
generated by revolving the region about the y-axis.
Solution: The axis of rotation is the y-axis and our variable of integration is x thus we
should use shellsa . Thus our volume is found by the integral
Z 1
1
x 2
dx
2π
ax + 1
0
a
if we wanted to use disk method we would need to integrate over y.
b) [8 pts] Evaluate your integral(s) in part a to find the volume of the solid.
Note: Your answer will possibly contain the constant a.
Solution: Let u = ax2 + 1. Then du = 2ax dx. At x = 0, u = 1 and at x = 1, u = a + 1.
Thus
Z 1
Z
1
π a+1 1
2π
x 2
dx =
du
ax + 1
a 1
u
0
a+1
π
=
ln |u|
a
1
π
π
=
ln |a + 1| − ln |1|
a
a
π
ln |a + 1|.
=
a