SAMPLE EXAM #1
Calculus II
SHOW ALL WORK – NO NEED TO SIMPLIFY ANSWERS
NO CALCULATORS
1. (20 points) Find the volume of rotation of the solid formed by revolving the
region bounded by f (x) = 2 − x2 and g(x) = 1 about the line y = −1 using the
disk method.
Solution:
Z
1
2 2
Z
1
π(3 − x ) dx −
volume =
−1
1
Z
=
π(22 ) dx
−1
π(9 − 6x2 + x4 ) dx − 8π
−1
1
1 5
3
= π 9x − 2x + x
− 8π
5
−1
1
1
− −9 + 2 −
− 8π
=π 9−2+
5
5
32π
=
.
5
2. (20 points) Find the volume of the solid described in the above problem using
the shell method.
Calculus 2
Exam #1, Page 2 of 6
Solution:
Z
2
volume =
2πρ(y)h(y) dy
1
Z
2
p
=
2π(1 + y) 2 2 − y dy
1
Z 2
p
= 4π
(1 + y) 2 − y dy
1
Calculus 2
Exam #1, Page 3 of 6
Let u = 2 − y. Then
du
dy
= −1 and dy = −du. Continuing
Z
♦
= 4π
√
(1 + y) u (−du)
♥
Z
♦
√
(3 − u) u du
= −4π
♥
Z ♦
since y = 2 − u
3u1/2 − u3/2 du
= −4π
♥
3/2
= −4π 2u
2
− u5/2
5
3/2
= −4π 2(2 − y)
2
= 0 − −4π 2 −
5
32π
=
.
5
♦
♥
2
− (2 − y)5/2
5
2
1
3. (20 points) Find the arclength of f (x) = 13 x3/2 for 0 ≤ x ≤ 60.
Calculus 2
Exam #1, Page 4 of 6
Solution:
Z bp
arclength =
1 + (f 0 (x))2 dx
a
s
2
Z 60
1 1/2
x
=
1+
dx
2
0
Z 60 r
x
=
1 + dx
4
0
Z ♣
√ du
=
u
letting u = 1 + x/4
1/4
♠
♣
2 3/2
=4
u
3
♠
60
8
3/2
= (1 + x/4)
3
0
8
3/2
3/2
16 − 1
=
= 168.
3
4. (20 points) Find the surface area generated by revolving the curve
√
3
y = 3x,
0≤y≤2
about the y–axis.
Solution: Notice that y =
√
3
3x
⇔
x = g(y) = 13 y 3 .
Calculus 2
Exam #1, Page 5 of 6
Z
b
p
2πρ(y) 1 + (g 0 (y))2 dy
Za 2 1 3 p
y
=
2π
1 + y 4 dy
3
0
Z 2 p
2π
y 3 1 + y 4 dy
=
3 0
Z
2π ♣ 3 √ du
y u 3
letting u = 1 + y 4
=
3 ♠
4y
Z
π ♣√
=
u du
6 ♠
♣
π 2 3/2
u
=
6 3
♠
2
π
(1 + y 4 )3/2 0
=
9
π
=
173/2 − 1 .
9
Surface Area =
5. (20 points) Find
0
a.) ln cos e3x
Z ln(6)
ex
dx.
b.)
ex + 4
0
Solution:
a.)
0
1
cos(e3x )
3x
cos(e )
0
1
=
− sin(e3x ) e3x
3x
cos(e )
−3e3x sin(e3x )
=
= −3e3x tan e3x .
3x
cos(e )
0
ln cos e3x
=
Calculus 2
Exam #1, Page 6 of 6
b.) Letting u = ex + 4,
Z
ln(6)
0
ex
dx =
ex + 4
Z
♣
♠
ex du
u ex
= [ln(u)]♣
♠
= [ln(ex + 4)]ln(6)
0
= ln(10) − ln(5) = ln(2).
EXTRA CREDIT
6. (20 points) The population of Normalville is 250 in 2005 and 350 in 2010. Assume the population of Normalville is increasing exponentially. When will the
population of Normalville reach 500?
Solution: Count time 2005 as year 0. Thus the initial population is Q0 =
250.
Q(t) = 250ekt
Q(5) = 350 = 250ek5
7
350
=
e5k =
250
5
5k = ln(7/5)
ln(7) − ln(5)
k=
5
One wants t such that
ln(7)−ln(5)
500 = 250e 5 t
ln(7)−ln(5)
2 = e 5 t
ln(7) − ln(5)
ln(2) =
t
5
5 ln(2)
t =
ln(7) − ln(5)
Thus the answer is 2005 +
5 ln(2)
ln(7)−ln(5)
which is 2015.3.