Engineering Math II – Spring 2015
Solutions for Class Activity #2
Problem 1. Find the area of the region bounded by the parabola y = x2 , the tangent line to
this parabola at (1, 1), and the x-axis. Then find the volume of the solid generated by rotating
this region about the x-axis.
Solution. Before we can even start to talk about area, we need to find the equation of the
tangent line to y = x2 at the point (1, 1). Since y 0 (x) is the slope of the tangent line to y at x,
and this line must pass through the point (1, 1), we can use the point-slope formula to find the
equation of the line. From the point-slope formula, we get
y − 1 = y 0 (1)(x − 1) = 2(x − 1)
since y 0 (x) = 2x. Solving the above equation for y yields
y = 2x − 1.
The following plot shows the graphs of y = x2 and y = 2x − 1, with the region bounded by
these curves and the x-axis shaded.
In order to find the area of the shaded region, we subtract the area under the “lower” curve
2
2
y = 2x − 1 from the area
R 1 2under the “upper” curve y = x . The area
R 1 under the curve y = x is
given by the integral 0 x dx, and area under 2x − 1 is given by 1/2 (2x − 1)dx. Thus the area
of the shaded region is given by
Z 1
Z 1
1
1 3 1 2
2
x dx −
(2x − 1)dx =
x − x −x 3 0
1/2
0
1/2
1
1 1 =
− 0 − (1 − 1) − ( − )
3
4 2
1 1
1
=
− = .
3 4
12
Solutions for Class Activity #2
2
Next, we find the volume of the solid obtained by rotating the same region about the x-axis.
In order to set up the integral, the first thing we need to determine is whether we will integrate
along the x-axis or the y-axis. It is often easiest to integrate along the same axis about which
the region is being rotated. In this case, that suggests we integrate along the x-axis.
The cross sections of the solid perpendicular to the x-axis are discs between x = 0 and
x = 12 , and are washers from x = 12 to x = 1. For this reason, we split our computation into
two parts. Between x = 0 and x = 21 , the radius of the disc is the distance from the x-axis to
the curve y = x2 , which is just x2 . Thus the volume is given by the integral
Z
1/2
Z
2 2
π(x ) dx = π
1/2
x4 dx
0
0
π 5 1/2
x 5
0
π 1
π
· 5 =
.
5 2
160
=
=
(1)
For the part from x = 21 to x = 1, the outer radius is the distance from the x-axis to the
curve y = x2 , which is just x2 , and the inner radius is the distance from the x-axis to the line
y = 2x − 1, which is just 2x − 1. So the volume of this piece is given by the integral
Z 1
Z 1
2 2
2
x4 − (4x2 − 4x + 1) dx
π(x ) − π(2x − 1) dx = π
1/2
1/2
1
Z
= π
(x4 − 4x2 + 4x − 1)dx
1/2
1
1
4
x5 − x3 + 2x2 − x 5
3
1/2
1 4
1
1 1 1 = π
− +1−
− + −
5 3
160 6 2 2
77 2
− −
= π −
15
480
13π
=
.
480
= π
The total volume is then given by the sum of the volumes in (1) and (2), which is
16π
π
480 = 40 .
π
160
(2)
+
13π
480
=
Integration along the y-axis.
If we did try to integrate along the y-axis, then the cross sections perpendicular to the y-axis
√
are discs. The radius of each disc, however, is half of the distance between the curve x = y
and the line x = y+1
2 , when y is between 0 and 1. Already one gets the feeling that integrating
along the y-axis is much more difficult to set up than integrating along the x-axis. As a rule of
thumb, it is often easiest to integrate along the axis perpendicular to the axis of rotation.
Solutions for Class Activity #2
3
Problem 2. Find the value(s) of c such that the area of the region bounded by the parabolas
y = x2 − c2 and y = c2 − x2 is 72.
Solution. The function y = x2 − c2 is a parabola opening upwards with vertex (0, −c2 ), and the
function y = c2 − x2 is a parabola opening downwards with vertex (0, c2 ). The case when c = 2
is shown in the following plot, with y = x2 − 4 in blue and y = 4 − x2 in red.
In the above plot, the two parabolas intersect each other when x = −2 and x = 2. In general,
the parabolas intersect each other when x2 − c2 = c2 − x2 . Solving for x gives x = ±c. Hence
the region bounded by the two parabolas is bounded on the x-axis between x = −c to x = c.
You may have noticed in the figure that the area of the region above the x-axis is equal to
the area below the x-axis. This is true for general c as well, since each parabola can be obtained
from the other by flipping it over the x-axis. Thus the area of the entire region is just twice the
area of the region above the x-axis. This area is easily computed by the integral
Z c
1 c
4c3
(c2 − x2 )dx = c2 x − x3 =
.
3
3
−c
−c
So the total area of the region is
c = 3.
8c3
3 .
If this area is 72, we have
8c3
3
= 72. Solving for c gives
Solutions for Class Activity #2
4
Problem 3. Sketch the region bounded by the given curves and find the area of the region:
4x + y 2 = 12,
x = y.
Solution. Solving the equation 4x + y 2 = 12 for x gives x = 3 − 41 y 2 , which is a parabola that
√
opens to the left with vertex (3, 0) and y-intercepts y = ± 12. This parabola, together with
the line x = y, are shown in the figure below.
The parabola intersects the line at the points (−6, −6) and (2, 2) (these points can be found
without graphing by solving 3 − 41 y 2 = y for y). It will be easiest to set up out integral over
the y-axis, since our functions are in terms of y. This will require splitting the computation up
into three parts representing the area of the regions A, B and C in the figure.
To compute the area of region A, we note that both curves are to the left of the y-axis, so
that the “top” curve is the line x = y, and the “bottom” curve is the parabola.√ The area is
computed by subtracting the area “under” the parabola from y = −6 to y = − 12 from the
area “under” the line from y = −6 to y = 0. The area under the line is given by
Z 0
Z −6
1 −6
−
ydy =
ydy = y 2 = 18.
2 0
−6
0
The area under the parabola is given by
Z −√12 Z −6 √
√
1 2
1 1 −6
−
3 − y dy = √
3 − y 2 dy = 3y − y 3 √ = 2 12 = 4 3.
4
4
12
− 12
−6
− 12
√
So the total area of region A is 18 − 4 3.
To compute the area of region B, first notice that the only curve bounding this region
(besides the axes) is the parabola. Since it is to the right of the y-axis, the area for this region
is given by
Z 0 √
√
1 1 0
3 − y 2 dy = 2y − y 3 √ = 12 = 2 3.
√
4
12
− 12
− 12
Solutions for Class Activity #2
5
Finally, to compute the area of region C, first note that the area√under the parabola from
y = 0 to y = 2 is the same as the area of region B, and so equal to 2 3. The area of region C
can be obtained by subtracting from this the area under the line x = y from y = 0 to y = 2.
This area is given by
Z 2
1 2
ydy = y 2 = 2.
2 0
0
√
Thus the area of region C is 2 3 − 2, which means that the total area is given by
√
√
√
area of A + area of B + area of C = 18 − 4 3 + 2 3 + 2 3 − 2 = 16.
Solutions for Class Activity #2
6
R π/2
Problem 4. Describe the solid represented by the integral π 0 [(1+cos(x))2 −12 ]dx. Evaluate
this integral. You will need to use the double-angle formula cos(2θ) = 2 cos2 (θ) − 1.
Solution. In the integral we see π out front and then the difference of two squares in the
integrand. This indicates some sort of rotation. The outer radius is the term 1 + cos(x), and
the inner radius is 1. Here is a plot of these two functions together.
The region being rotated is from x = 0 to x = π/2, which is the region bounded by these two
curves to the right of the y-axis. When rotated about the x-axis, this gives a rounded washer
with a flat base. To evaluate the integral, we first simplify the integrand and split the integral
into two integrals:
Z π/2
Z π/2
2
2
(1 + cos(x)) − 1 dx =
cos2 (x) + 2 cos(x) dx
0
0
Z
=
π/2
2
Z
cos (x)dx + 2
0
This second integral is rather easy to evaluate:
Z π/2
π/2
2
cos(x)dx = 2 sin(x)0 = 2.
π/2
cos(x)dx.
0
(3)
0
For the other integral, we need to use the double-angle formula and replace cos2 (x) with
cos(2x)+1
.
2
Z π/2
Z
1 π/2
2
cos (x)dx =
cos(2x) + 1 dx
2 0
0
1
1 π/2
=
sin(2x) + x 4
2
0
π
=
.
(4)
4
So the value of the original integral is the sum of the values in equations (3) and (4); that is,
2 + π4 .
Solutions for Class Activity #2
7
Problem 5. Find the volume of the solid obtained by rotating the region bounded by the
curves y = 1 + sec(x) and y = 3 about the line y = 1.
Solution. First we plot the relevant functions. In the following graph, 1 + sec(x) is in blue,
y = 3 in red, and y = 1 in yellow.
The region is bounded on the x-axis between x = − π3 and x = π3 . When rotated about the
line y = 1, cross sections perpendicular to the x-axis are washers with outer radius 2 and inner
radius sec(x). Thus the volume is given by the integral
Z
π/3
π
−π/3
π/3
22 − sec2 (x) dx = π 4x − tan(x) −π/3
=
=
4π 2 √
− 3π −
3
√
8π 2
− 2 3π
3
−
4π 2 √ + 3π
3
Solutions for Class Activity #2
8
Problem 6. Find the volume of the solid obtained by rotating the region bounded by the
curves y = x2 and x = y 2 about the line x = −1.
Solution. The function y = x2 is a parabola opening upwards with vertex at the origin, and
the function x = y 2 is a parabola opening to the right, also with vertex at the origin. See the
figure blow.
The region bounded by the these parabolas is bounded on the y-axis between y = 0 and y = 1
(since the parabolas intersect at the origin and the point (1, 1)). When rotated about the
vertical line x = −1, cross sections perpendicular to the y-axis are washers. In order to find
the outer radius, we first note that since the region lies to the right of the y-axis, we only care
about the portion of the parabola y = x2 corresponding to positive x values, and so rewrite its
√
√
equation as x = y. Then the outer radius is 1 + y, and the inner radius is 1 + y 2 . Thus the
volume of the solid is given by the integral
Z 1
Z 1
√ 2
√
2 2
π
(1 + y) − (1 + y ) dy = π
1 + 2 y + y − (1 + 2y 2 + y 4 ) dy
0
0
Z 1
√
= π
(y + 2 y − 2y 2 − y 4 )dy
0
1
4 3 2
1 1
= π y2 + y 2 − y3 − y5 2
3
3
5
0
1 4 2 1
= π
+ − −
2 3 3 5
29π
=
.
30
Solutions for Class Activity #2
9
√
Problem 7. Let R be the region bounded by the y-axis, the line y = 1, and the curve y = x.
Compute the volume of the solid whose base is R and whose cross sections perpendicular to the
y-axis are semicircles.
Solution. We need to find the area of a a cross section as a function of either x or y. Since the
cross sections are perpendicular to the y-axis, it makes the most sense to express the area of
each cross section as a function of y. In order to find an expression for this area, we first plot
the given functions to find the region R.
Since the cross sections are semicircles, the area is given by 12 πr2 , where r is the radius of the
semicircle. The diameter of the semicircle lies between the y-axis and the graph of the function
√
y = x, or x = y 2 when written as a function of y. Thus the radius of a cross section is given
by 21 y 2 , and the integral for determining the volume of the solid is
π
2
Z
0
1 2 2
y
2
dy =
=
=
Z
π 1 4
y dy
8 0
π 1 5 1
y 8 5
0
π
.
40
Solutions for Class Activity #2
10
Problem 8. A hole of radius r is bored through the center of a sphere of radius R > r. Find
the volume of the remaining portion of the sphere. [Recall that the volume of a sphere with
radius R is 43 πR3 .]