Download Exam 3 Solutions

Math 142.009
Exam 3 Solutions
NAME:
#1.)
/10
#5.)
/10
#2.)
/10
#6.)
#3.)
/15
/15
#7.)
/20
#4.)
Total:
/20
/100
Instructions: There are 5 pages and a total of 100 points on the exam. You must show all necessary
work to get credit. You may not use your book, notes, or calculator. Partial credit will only be given for
progress toward a correct solution. Once you get your answer down to numbers only, you do not have to
simplify.
1.) (10 points) Find the area of the region in the first quadrant enclosed by the curves y = 4 − x2 , y = 3x,
and the y-axis. (The first quadrant is the region where x > 0 and y > 0.)
y = 4 − x2
y
(1, 3)
y = 3x
x
The intersection of the curves occurs when 4 − x2 = 3x, or x2 + 3x − 4 = 0, or (x + 4)(x − 1) = 0,
hence x = −4 or x = +1. Since we are only interested in the first quadrant, x = +1, and the intersection
point is (1, 3). To find the area, it is easier to slice the region vertically, since each vertical slice runs from
y = 3x to y = 4 − x2 , and hence has length `(x) = 4 − x2 − 3x. We have one of these slices for every x
between x = 0 and x = 1. So the area is
A=
Z
1
`(x) dx =
0
Z
1
24 − 2 − 9
13
1 3
1 3 3 2 1
= .
4 − x − 3x dx = 4x − x − x = 4 − − − 0 =
3
2
3 2
6
6
0
2
0
2.) (10 points) Find the length of the curve y = 2x3/2 from x = 1 to x = 4.
√
dy
= 3x1/2 = 3 x,
Zdx4 √
2
2
dy
dy
so dx = 9x. Hence 1 + dx = 1 + 9x. Therefore the length of the curve is ` =
1 + 9x dx. We
1
evaluate this integral using the substitution u = 1 + 9x. Then du = 9 dx. When x = 1, we have u = 10,
and when x = 4, we have u = 37. Hence
By the arc length formula, the length of the curve is
`=
Z
37
10
R4q
1
1 + (dy/dx)2 dx. We get
√ 37
1 2
2 3/2
2 √
1 1/2
u du = · u3/2 =
37 − 103/2 =
37 37 − 10 10 .
9
9 3
27
27
10
1
3.) (15 points) The region in the first quadrant enclosed by the curves y = x3 and y = 4x is revolved
around the x-axis. Find the volume of the resulting solid.
y
(2, 8)
y = 4x
y = x3
x
To find the intersection of the line y = 4x and the curve y = x3 , we set 4x = x3 to get x3 − 4x = 0.
We solve this by factoring: x3 − 4x = 0 or x(x2 − 4) = 0 or x(x − 2)(x + 2) = 0, so either x = 0, x = −2,
orx = +2. Since we are looking only at the first quadrant, we have x = 0 or x = 2; substituting into
either equation gives y = 0 when x = 0 and y = 8 when x = 2, so the intersection points are (0, 0) and
(2, 8). Since we are revolving the region around the x-axis, we slice this region vertically. We have one of
these slices for every value of x from x = 0 to x = 2. For a fixed x between x = 0 and x = 2, the slice
revolved around the x-axis yields a washer with area π(R2 − r2 ). Here the outer radius R of the washer
is the y-value on the graph y = 4x, so R = 4x. The inner radius r is the y-value on the graph y = x3 , so
r = x3 . Hence the area of the slice at x is A(x) = π((4x)2 − (x3 )2 ) = π(16x2 − x6 ). Therefore the volume
is
Z 2
Z 2
16
1 7
16 3 1 7 2
2
6
x − x =π
·8− ·2 −0
V =
A(x) dx =
π(16x − x ) dx = π
3
7
3
7
0
0
0
4
512
1 1
7−3
128 128
−
−
= 128π ·
=
π.
= 128π
= 128π ·
=π
3
7
3 7
21
21
21
2
4.) (20 points) A water tank is is in the shape of a cone with the vertex (point) on the ground. The tank
has radius 5 ft and height 10 ft. The tank is full of water which weighs 62.5 lbs/ft3 . Find the work done, in
ft-lbs, in pumping all the water out through a point at the top of the tank. Recall that once your answer
only has numbers in it, you can stop. Constants like π and e count as numbers in this regard.
5
y
R
10 − y
+y
We start by setting up a coordinate line with origin at the center of the top of the cone, with positive
y-axis going down. To find the total work, we will look at the water in a slice of width dy at depth y
and compute the work done in pumping that slice up to the opening. To get the total work done, we
will add up, via an integral, the work for these slices for every y starting at y = 0 and ending at y = 10.
At a particular depth y, the work done in lifting the slice is the weight times the distance. The distance
the water at height y must go is d = y, by our choice of coordinates. The weight of the slice of water at
depth y is the volume V (y) times the density ρ = 62.5, so the weight is ρV (y). The volume of the slice
with thickness dy is V (y) = A(y)dy, where A(y) is the area of the cross-section at depth y. Since the slice
is a disk, its area is πR2 , where R is the radius of the disk. To find R, we use similar triangles to say
R
5
1
1
1
that
=
= , so we get R = (10 − y) = 5 − y. Putting this back together, the area A(y) is
10 − y
10
2
2
2
2
1
1 2
2
πR = π 5 − y . So the volume at height y is V (y) = A(y) dy = π 5 − y dy. Hence the weight at
2
2
1 2
height y is ρV (y) = ρπ 5 − y dy. Multiplying by the distance y that the water at depth y is raised,
2
we get that the work in raising the water at depth y to the top is
1
ρπ 5 − y
2
2
1
1
y dy = ρπ 25 − 5y + y 2 y dy = ρπ 25y − 5y 2 + y 3 dy.
4
4
Hence the total work done in pumping all the water out is
W =
Z
10
0
1
ρπ 25y − 5y + y 3 dy = ρπ
4
= 2, 500ρπ
2
2, 500 5, 000 10, 000
25 2 5 3
1 4 10
y − y + y = ρπ
−
+
2
3
16
2
3
16
0
1 2 1
6−8+3
1
− +
= 2, 500ρπ
= 2, 500(62.5)π
≈ 40, 906 ft-lbs.
2 3 4
12
12
3
5.) (10 points) Find the average value of the function f (x) = sin2 x on the interval [π, 2π].
1 Zb
f (x) dx for the average value of a function f on an interval [a, b], in this case
b − a aZ
1 2π 2
the average value is A =
sin x dx. We evaluate this integral using the identity sin2 x = 21 − 12 cos(2x).
π π
Hence
1 Z 2π 1 1
A=
− cos(2x) dx
π π 2 2
2π
1
1 1
x − sin(2x) =
π 2
4
π
1
1
1
1 1
(2π) − sin(4π) −
π − sin(2π)
=
π 2
4
2
4
π
1 π
1
1
π−
= · = ,
=
π
2
π 2
2
since sin(2π) = 0 and sin(4π) = 0
By the formula
6.) (15
√ points) Find the centroid (x, y) of the region in the first quadrant bounded above by the curve
y = x, below by the x-axis, and on the right by x = 4.
Rb
xf (x) dx
We use the formulas x = R b
and y =
a f (x) dx
a
y
y=
√
x
(4, 2)
.
x=4
Rb
√
(f (x))2 dx
, where f (x) = x, a = 0, and b = 4.
a f (x) dx
1
a 2
Rb
x
We calculate
Z
Z
b
a
b
a
f (x) dx =
xf (x) dx =
and
Z
b
a
Hence
Z
Z
4
0
4
0
√
x dx =
Z
4
x
1/2
dx =
0
2 3/2 4 2 3/2
2
16
x
− 0) = · 8 = ,
= (4
3
3
3
3
0
Z 4
√
2
64
2 5/2 4 2 5/2
x
− 0) = · 32 = ,
x3/2 dx =
x x dx =
= (4
5
5
5
5
0
0
Z 4 √ 2
1Z 4
1
1
(f (x))2 dx =
x dx =
x dx =
2
2 0
0 2
1 2 4 16
x =
− 0 = 4.
4
4
0
64/5
64 3
12
4
3
3
=
·
=
and y =
=4·
= .
16/3
5 16
5
16/3
16
4
12 3
Hence the centroid of the region is (x, y) =
,
= (2.4, 0.75).
5 4
x=
4
7.) (20 points) The region enclosed by the curves y = x2 , x = 2 and the x-axis is revolved around the line
x = 2. Find the volume of the resulting solid.
y
(2, 4)
y = x2
√
x= y
y
x
x=2
For the curve y = x2 , we have y = 4 when x = 2. Since we are revolving around the line x = 2,
we slice this region horizontally. We have one of these slices for every value of y from y = 0 to y = 4.
For a fixed y between y = 0 and y = 4, the slice revolved around the line x = 2 yields a disk with
area πR2 . Here the radius R of the disk is the x-value on the graph x = 2, minus the x-value on
√
√
the graph y = x2 , or x = y. That is, R = 2 − y. Hence the area of the slice at height y is
√
√
A(y) = π(2 − y)2 = π(4 − 4 y + y) = π(4 − 4y 1/2 + y). Therefore the volume of the solid is
V =
y2
2
= π 4y − 4 · y 3/2 +
3
2
Z
4
A(y) dy =
0
Z
4
0
π 4 − 4y 1/2 + y dy
!4
64
72 − 64
8
8
=π·
= π.
= π 16 − · 8 + 8 = π · 24 −
3
3
3
3
0
5