Math 126
Name:
Summer 2014
Score:
/33
Show all your work
Dr. Lily Yen
No Calculator permitted in this part. Read the questions carefully. Show all your work
and clearly indicate your final answer. Use proper notation.
Problem 1: First draw y = x2 + 4 and y = 10. Shade the region in the first quadrant
bounded by these two graphs and the y-axis.
Test 2
y
√
14
12 − y
y−4
12
10
8
6
3
3+
−8
−7
−6
−5
−4
−3
−2
√
4
y 2− 4
−1
1
2
x
Use integrals to express the following. Do not evaluate your integrals. Draw a
cross-sectional strip for each solid of rotation.
a. The area of the shaded region.
If x2 + 4 = 10 and x ≥ 0, then x =
√
√
6. Therefore the area is
√
6
√
Score:
/2
10 − (x + 4) dx =
6 − x2 dx
=4 6
0
0
b. The volume obtained
if the region is rotated
about x = −3 , using the method of
slicing (washers/disks).
Z 10 p
( y − 4 + 3)2 − 32 dy
π
Z
6
Z
2
4
10
√
= π( 12 y 2 − 4y + 4(y − 4)3/2 )4 = π(18 + 24 6)
Score:
/2
c. The volume obtained if the region is rotated about y = 12, using the method of
cylindrical shells.
Z 10
p
2π
(12 − y) y − 4 dy
4
10
(y − 4)3/2 )4 =
= π(− 25 (y − 4)5/2 + 16
3
176
π
5
√
6
Score:
/2
d. The volume of a solid that has the shaded region as its base, and cross-sections perpendicular to the x-axis are squares.
√
Z
√
6
(10 − (x2 + 4))2 dx =
0
√6
= 15 x5 − 4x3 + 36x0 =
Z
6
(6 − x2 )2 dx
0
√
96
5
6
Score:
/2
e. The volume of a solid with the shaded region as its base, and cross-sections perpendicular to the y-axis are equilateral triangles.
Z 10 √ p
Z 10 √
3
3
2
( y − 4) dy =
(y − 4) dy
4
4
4
4
Score:
/2
Problem 2: First, draw the following graphs
on the coordinate system, then shade the region bounded by them:
f (x) =
8
,
x2
g(x) = 8x,
and h(x) = x.
y
g
8
6
4
h
2
f
Set up the integral for the area of the
shaded region, then evaluate the integral.
1
x
2
If f (x) = g(x), then x82 = 8x, so 1 = x3 , so x = 1. If f (x) = h(x), then x82 = x, so 8 = x3 , so
x = 2. If g(x) = h(x), then 8x = x, so x = 0. Thus the intersection points are located as
indicated in the graph.
The area is then
Z 2
Z 1
Z 2
Z 1
8
8x − x dx +
f (x) − h(x) dx =
g(x) − h(x) dx +
− x dx
2
1 x
0
1
0
Z 2
Z 1
8x−2 − x dx
7x dx +
=
1
0
2
7 2 1
= x + (−8)x−1 − 1 x2 2
7
2
0
8
2
= − −2+8+
=6
2
1
1
2
Score:
/4
Problem 3: The solid S is the intersection of two cylinders of radius r whose axes are
perpendicular. Find the volume of S as a function of r.
y
p
The dashed rectangle in the figure would have width 2 r2 − y 2 . The shown dashed
rectangle intersects
with a similar rectangle inside the other cylinder creating a square with
p
2
side length 2 r − y 2 . The solid S is formed by stacking all these squares, so the volume
of S is
Z r
r
2
2
2
4 3
4(r − y ) dy = 4r y − 3 y = 16
r3
3
−r
−r
Score:
Page 2
/4
Math 126
Math 126
Name:
Summer 2014
Show all your work
Dr. Lily Yen
Calculators permitted from here on.
Problem 4: The acceleration of a Nerf bullet is a(t) = 60t − 4t3 meters per second squared
after it is fired. Compute the average acceleration and the average velocity over the time
interval [2, 6], assuming that the particle’s initial velocity is zero (that is, v(0) = 0).
R6
R
1 6
1
1
3
2
4 6
a(t)
dt
=
60t
−
4t
dt
=
(30t
−
t
)
= −80 m/s2 .
The average acceleration is 6−2
4 2 R
4
2 R
2
t
t
t
The velocity of the bullet is v(t) = 0 a(u) du = 0 60u − 4u3 du = 30u2 − u4 |0 = 30t2 − t4 .
Therefore
the average
velocity is
R6
R
1 6
1
1 5 6
1
2
4
3
v(t)
dt
=
30t
−
t
dt
=
= 132.8 m/s
(10t
−
t
)
6−2 2
4 2
4
5
2
Test 2
Score:
/4
Problem 5: If 5 J of work are needed to stretch a spring 10 cm beyond equilibrium, how
much work is required to stretch it 15 cm beyond equilibrium?
Work is force times distance, E = F × d, and the force of a spring is F = k × d, where k is
the spring constant. Thus the work required to stretch a spring a distance of d beyond its
Rd
Rd
d
equilibrium is E = 0 F dx = 0 kx dx = 12 kx2 0 = 21 kd2 .
In standard units, 10 cm = 0.10 m, so 5 J = 21 k(0.10 m)2 , so
2×5
2
k = 0.10
2 = 1000 J/m = 1000 N/m.
Therefore the work required to stretch the spring 15 cm = 0.15 m is
E = 12 × 1000 N/m × (0.15 m)2 = 11.25 J.
Score:
Page 3
/4
Math 126
Problem 6: Calculate the work required to lift a 10 m chain over the side of a building.
Assume that the chain has a density of 8 kg/m.
The work required to list a mass m the distance x is mgx, where g is the acceleration due
to gravity.
The short piece of chain with length dy at height y has mass 8 dy and needs to be lifted a
distance of 10 − y. Thus the work required to lift the chain is
Z 10
1
8g(10 − y) dy = 8g(10y − 21 y 2 )0 0 = 400g = 3920 J
0
using 9.8 m/s2 for g.
Score:
/3
Problem 7: Water is pumped into a spherical tank of radius 2 m from a source located 1 m
below a hole at the bottom. The density of water is 1000 kg/m3 . Calculate the work F (h)
required to fill the tank to level h metres in the sphere.
y
2m
2
r
1m
Water source
The p
cross section of the water at a distance y from the centre of the tank has radius
r = 22 − y 2 . Hence the slab of water of thickness dy has volume πr2 dy = π(4 − y 2 ) dy
and mass 1000π(4 − y 2 ) dy. Since the slab has to be lifted a distance of 1 + 2 + y, the work
required for the slab is 1000π(4 − y 2 )(3 + y)g dy = 1000π(12 + 4y − 3y 2 − y 3 )g dy, where
g = 9.8 m/s2 is the acceleration due to gravity. Thus
Z
h−2
F (h) =
1000π(12 + 4y − 3y 2 − y 3 )g dy
−2
h−2
= 1000π(12y + 2y 2 − y 3 − 41 y 4 )g −2
= 1000π(−12 + 2h2 + h3 − 41 h4 − (−12))g
= 1000π(2h2 + h3 − 14 h4 )g
(in joule, J).
Score:
Page 4
/4
Math 126