Math 21B
1 (20 pts.)
Brian Osserman
Practice Exam 2
The region enclosed by the curves
y = x1/2 ,
y = x1/4 ,
0≤x≤1
is revolved about the y -axis. Find the volume of the resulting solid.
Solution:
We use cylindrical shells. For 0 ≤ x ≤ 1 we have x1/4 ≥ x1/2 ,
so the height of a shell is given by x1/4 − x1/2 . Since we revolve about the
y -axis, the radius is x, so the volume is
Z 1
Z 1
1/4
1/2
2π
x(x − x )dx = 2π
(x5/4 − x3/2 )dx
0
0
1 !
4 9/4 2 5/2
= 2π
x − x
9
5
0
4 2
= 2π
− −0
9 5
4
=
π.
45
2 (20 pts.)
The parametric curve
2
x = t3/2 ,
3
√
y = 2 t,
0≤t≤
√
3
is revolved about the y -axis. Find the area of the resulting surface.
Solution:
The area is
√
Z
s
3
x
2π
0
dx 2 dy 2
+
dt = 2π
dt
dt
√
Z
0
√
3
2 3/2
t
3
q
(t1/2 )2 + (t−1/2 )2 dt
2 3/2 √
t
t + t−1 dt
3
0
Z √3 √
4
=
π
t4 + t2 dt
3 0
Z √3 √
4
=
π
t t2 + 1dt.
3 0
Z
= 2π
3
Using u = t2 + 1, du = 2tdt, this gives
2
π
3
Z
4
√
1
2
π
udu =
3
2 3/2
u
3
4 !
1
2
2
2
=
π
·8− ·1
3
3
3
14
2
π·
=
3
3
28
=
π.
9
3 (15 pts.)
If a force of 90 N stretches a spring 1 m beyond its natural length, how
much work does it take to stretch the spring 5 m beyond its natural length?
Solution:
If the force constant for the spring is k N/m, we have k · 1 = 90,
so k = 90. Then the work to stretch the spring 5 m is (in Newtons):
Z
5
90xdx = 45x2
5
0
0
= 45 · 25 − 0
= 1125.
4 (20 pts.)
The shape of a water tank is determined by revolving the curve y = x2 ,
0 ≤ x ≤ 4 about the y -axis. The container is lled with water. What is
the work required to pump all the water out over the top edge of the tank?
(the weight-density of water is 10, 000 N/m3 , and assume all distance units
are in meters).
Solution:
If we take a horizontal slice of water at height y , with thickness
dy , we get a disk of radius x =
√
y . Since x goes from 0 to 4, we see that y
goes from 0 to 16, so the tank has height 16, and we need to lift this slice
a distance of 16 − y m. The weight of the slice in Newtons is
√
10, 000 · π( y)2 dy = 10, 000πydy.
Thus, the work to pump out the slice is 10, 000πy(16 − y)dy N·m. To nd
the total work to pump the water out, we have to integrate over all slices,
2
so we get (in N·m):
16
Z
1
8y 2 − y 3
3
10, 000πy(16 − y)dy = 10, 000π
0
16 !
0
4096
= 10, 000π 2048 −
−0
3
20, 480, 000
=
π.
3
5 (15 pts.)
A cubical tank with sides 3 m long is lled with water to a depth of 2 m.
Find the total force of the water on one side of the tank (the weight-density
of water is 10, 000 N/m3 ).
Solution:
At a depth of h meters, the pressure of the water on the side is
10, 000h N/m2 . The length of a side is 3 m, so (in Newtons) the total force
is
Z
2
Z
10, 000h · 3dh = 30, 000
0
2
hdh
0
= 30, 000
1 2
h
2
2 !
0
= 30, 000(2 − 0)
= 60, 000.
6 (10 pts.)
Evaluate the following integral
Z
(csc x − sec x)(sin x + cos x)dx.
Solution:
Z
(csc x − sec x)(sin x + cos x)dx
Z
(csc x sin x − sec x cos x + csc x cos x − sec x sin x)dx
=
Z
=
(1 − 1 + cot x − tan x)dx
= ln | sin x| + ln | cos x| + C.
3