Exam 2 Review Solutions
Math 2502: Spring 2013
√
1. Let R be the regions bounded by the graphs of y = x and y = 2 x. Set up, but do not evaluate, the
integral which computes the volume of the solid obtained by rotating R
(a)
(b)
(c)
(d)
about
about
about
about
the
the
the
the
x-axis.
y-axis.
line x = −1.
line y = 5.
Solution: There are a few ways of doing this problem, since the region can be easily cut into either
horizontal or vertical box cross-sections. However, I will be using the vertical box cross-sections
since all expressions are in terms of x. To find the bounds of x, we set the two functions equal to
each other:
√
2 x=x
4x = x2
0 = x2 − 4x
0 = x(x − 4)
so that x varies on the interval [0, 4]. (a) When the cross-section shown below is rotated around the
x-axis one gets a “washer.”
√
y=2 x
y
y=x
R
r
x
The volume of this washer is
dV = (πR2 − πr2 ) dx
√
= (π(2 x)2 − π(x)2 ) dx
Z 4
√
⇒V =
(π(2 x)2 − π(x)2 ) dx
0
1
(b) When the cross-section is rotated around the y-axis one gets a “shell.”
√
y=2 x
y
y=x
r
h
x
The volume of this shell is
dV = 2πrh dx
√
= 2πx(2 x − x) dx
Z 4
√
⇒ V = 2π
x(2 x − x) dx
0
(c) When the cross-section is rotated around the line x = −1 one gets a “shell.”
√
y=2 x
y
y=x
r
h
x
The volume of this shell is
dV = 2πrh dx
√
= 2π(x + 1)(2 x − x) dx
Z 4
√
⇒ V = 2π
(x + 1)(2 x − x) dx
0
Page 2
(d) When the cross-section shown below is rotated around the line y = 5 one gets a “washer.”
y
√
y=2 x
r
y=x
R
x
The volume of this washer is
dV = (πR2 − πr2 ) dx
√
= (π(5 − x)2 − π(5 − 2 x)2 ) dx
Z 4
√
⇒V =
(π(5 − x)2 − π(5 − 2 x)2 ) dx
0
2. Evaluate the following indefinite integrals.
R
(a) e3x cos x dx
R
(b) sin5 x cos x dx
R
(c) tan5 x sec3 x dx
Solution: (a) This is the product of two functions, so we will implement the method of integration
by parts. Let u = e3x and dv = cos x dx, so that du = 3e3x dx and v = sin x.
Z
Z
Z
e3x cos x dx = uv − v du = e3x sin x − 3e3x sin x dx
Page 3
Now let u = 3e3x and dv = sin x dx so that du = 9e3x dx and v = − cos x
Z
Z
e3x cos x dx = e3x sin x − uv − v du
Z
Z
e3x cos x dx = e3x sin x − −3e3x cos x + 9e3x cos x dx
Z
Z
e3x cos x dx = e3x sin x + 3e3x cos x − 9 e3x cos x dx
Z
10 e3x cos x dx = e3x sin x + 3e3x cos x
Z
1 3x
e3x cos x dx =
e sin x + 3e3x cos x + C
10
(b) Let u = sin x, so that du = cos x dx
Z
Z
5
sin x cos x dx = u5 du
u6
+C
6
1
= sin6 x + C
6
=
(c) If we save sec x tan x, then u = sec x, and even powers of tan x can be changed into sec x via
tan2 x = sec2 x − 1.
Z
tan5 x sec3 x dx =
Z
Z
=
Z
=
Z
=
tan4 x sec2 x(sec x tan x dx)
(sec2 x − 1)2 sec2 x(sec x tan x dx)
(u2 − 1)2 u2 du
(u4 − 2u2 + 1)u2 du
Z
(u6 − 2u4 + u2 ) du
5
u7
u
u3
=
−2
+
+C
7
5
3
1
2
1
= sec7 x − sec5 x + sec3 x + C
7
5
3
=
3. Find the arclength of the graph of y = x2 − x + 1 on the interval [0, 1]. Hint. Use trigonometric
substitution.
Page 4
Solution: The arclength formula is
Z
b
p
s=
1 + (dy/dx)2 dx
a
Z
s=
1
p
1 + (2x − 1)2 dx
0
Let tan θ = 2x − 1 so that
d
d
tan θ =
(2x − 1)
dθ
dθ
dx
sec2 θ = 2
dθ
1
⇒ dx = sec2 θ dθ
2
Plugging these back into the integral:
Z p
Z p
1
1 + (2x − 1)2 dx =
1 + tan2 θ
sec2 θ dθ
2
Z
1 √ 2
sec θ sec2 θ dθ
=
2
Z
1
=
sec3 θ dθ
2
Using the reduction formula for secant:
Z
Z
1
1 sec θ tan θ 1
1
1
3
sec θ dθ =
+
sec θ dθ = sec θ tan θ + ln | sec θ + tan θ |
2
2
2
2
4
4
To get the expression in terms of x, we draw a right triangle and use trigonometry. Since tan θ =
2x − 1 = opp/adj, we can fill in the rest of the right triangle with the Pythagorean theorem:
p
1 + (2x − 1)2
2x − 1
1
In particular, sec θ =
Z
s=
p
1 + (2x − 1)2 and
1
p
1 + (2x − 1)2 dx
0
1
p
p
1
1
= (2x − 1) 1 + (2x − 1)2 + ln | 1 + (2x − 1)2 + (2x − 1) |
4
4
0
p
p
√
1
1
1
1 √
(1) 1 + 1 + ln | 1 + 1 + 1 | −
(−1) 1 + (−1)2 + ln | 1 + (−1)2 + (−1) |
=
4
4
4
4
√
√
√
2 1
1
=
+ ln( 2 + 1) − ln( 2 − 1)
2
4
4
Page 5
4. A water trough has a cross section equal to the region enclosed by y = x2 and y = 1, and a length of 3
meters. It is filled with water.
(a) How much work is required to pump the water out of the trough from the top? The density of
3
water is ρ = 1000 kg/m .
(b) Find the total force on the face of the trough.
Solution: (a) First draw the picture.
y
y
x
x
The small work dW done on this cross section of water is force·distance. The force is the weight
(ρgdV , gravity is included since in metric units) of the cross section, and the distance is the distance
(1 − y) one needs to lift the cross section to the top of the trough. The volume dV = `wh =
3(2x)dy = 6xdy. Thus,
dW = 6ρgx(1 − y)dy
√
Since y = x2 provides the relation between x and y for the width of the cross section, x = y and
Z
W =
0
1
24
√
6ρg y(1 − y) dy =
ρg = 15680J
15
(b) First draw the picture.
y
y
x
x
Page 6
Hydrostatic pressure is given by P = ρgh, where h is the depth of the water. The total force is
F = P A, where A is the area on which the pressure acts. For the given cross section h = 1 − y and
A = 2xdy so that the force on the face of the trough on this section is
dF = 2ρgx(1 − y)dy
In particular, x =
√
y and
Z
F =
0
1
8
√
2ρg y(1 − y)dy =
ρg ≈ 5227N
15
Page 7