Davis Buenger
Math 1172 6.4
September 1, 2015
1. (Test Question) True or False: Given a solid generated by revolving a region about the
line x = 3, if we are using the shell method to compute its volume, then we would integrate
with respect to x.
True: The line x = 3 is parallel to the y-axis. When you use shells the variable of integration
is opposite to the axis of rotation, so we should integrate over x if we want to use the shell
method
2. Let R be the region in the first quadrant bounded by x = y 2 , y=0, and x = 4. Use the
shell method to find the volume of the solid generated when R is revolved around the x-axis.1
Solution: As noted by the problem we will
use the shell method to find the volume of
this shape. Since we are rotating around the
x−axis, in order to use the shell method we
should integrate over y.
y value 2 and our region is bounded from below by the line y = 0.a Thus our interval of
integration is [0, 2]. The top function is x = 4
and our bottom function is x = y 2 .b Hence
Z
2
V ol = 2π
y(4 − y 2 ) dy
Z0 2
4y − y 3 dy
0
2
2
y
y4
= 2π 4 −
2
4 0
2
= 2π 2(2 ) − 24 /4 − 0
= 8π.
= 2π
The curves x = 4 and x = y 2 intersect at the
a
Notice that the curves actually enclose two regions, one below the x−axis and one above. When rotated
around the x−axis, each region carves out the same shape. Hence we could apply the shell method to either
one, but we should not apply it to both as that will double the region and the volume.
b
Please excuse the ambiguous use of the words top and bottom. Recall that we consider our shape in
terms of y. So the function x = 4 is on top of the function x = y 2 for y in the interval [0, 2] because for any
y value in that interval 4 ≥ y 2 .
3. (Test Question) A solid S is generated by rotating the region bounded by y = sin(x) and
y = cos(x) for values of 0 ≤ x ≤ π/4 around the line x = −1. Find an integral representing
the volume of S.
1
This problem is particularly suited for the shell method since we are rotating around the x−axis and
our functions are mainly functions of y.
1
Solution: As our functions are given as functions of x and the line x = −1 is parallel to
the y-axis, the instinctive choice for finding
the volume is the shell method. On the interval [0, π/4],
√ sin(x) and cos(x) intersect only
at (π/4, 2/2). Combining this with the fact
that cos(0) = 1 > 0 = sin(0), it follows that
cos(x) ≥ sin(x) on [0, π/4]. Hence the height
of our shells is given by
For a fixed x0 , the distance from (x0 , 0) to the
line x = −1 is x0 + 1. Hence our radius as a
function of x is given by the expression
r(x) = x + 1.
Thus
Z
π/4
(x + 1)(cos(x) − sin(x)).
V ol = 2π
0
h(x) = cos(x) − sin(x).
4. Let R be the region bounded by y = x − x4 and y = 0. Let S be the solid generated by
revolving R about the y-axis. Use either washers method or shell method to find the volume
of S.
Solution: Our given functions are in terms
of x and we are rotating around the y−axis,
thus we should use shell method. A general
formula for the shell method in this case looks
like
Z b
x(top(x) − bot(x))dx
2π
a
Setting y = x − x4 and y = 0 equal we find
a
that the two curves intersect at 0 and 1:
0 = x − x4 = x(1 − x)(1 + x + x2 ).a
On the interval [0, 1] y = x − x4 is the top
function and y = 0 is the bottom function.
Thus
Z 1
V ol = 2π
x(x − x4 − 0) dx
0
Z 1
= 2π
x2 − x5 dx
0
3
1
x
x6
= 2π
−
3
6 0
3
3
1
16
0
06
= 2π
−
− 2π
−
3
6
3
6
= π.
1 + x + x2 has no real roots as seen from the quadratic formula
2
√
−1± 12 −4∗1∗1
.
2
5. Let R be the region bounded by y = 6/(x + 3) and y = 2 − x. Let S be the solid generated
by revolving R about the x-axis. Find the volume of S first by using washers method and
second by using shell method. Which is easier?
Solution: Let us calculate the intersection Now let us try the shell method. Let us solve
of these two curves:
our equations to be functions of y. The sec6
ond is x = 2 − y and the first:
= 2−x
x+3
y(x + 3) = 6
6 = 6 − x − x2
yx + 3y = 6
0 = −x2 − x
6 − 3y
0 = x(x + 1).
x =
y
Thus the curves intersect at the points
6
=
− 3.
(0, 2) and (−1, 3). Let us start with the
y
washer/disk method.
Since we are rotating around the x-axis, we By evaluating each function at y = 52 , we see
will need to integrate with respect to x to use that x = 2 − y is the top function on the
the washer/disk method.
y-interval [2, 3]. Hence by the shell method:
−1
By evaluating the functions at x = 2 , one
Z 3
6
sees that y = 2 − x is the top function on the
y(2 − y − ( − 3)) dy
V olume = 2π
x-interval [−1, 0]. Hence by the washer/disk
y
Z2 3
method:
2y − y 2 − 6 + 3y dy
= 2π
V olume
2
Z 0
Z2 3
6
= π
(2 − x)2 −
dx
5y − y 2 − 6 dy
= 2π
3
+
x
−1
2
Z 0
3
36
5 2 1 3
2
dx
= π
4 − 4x + x −
= 2π y − y − 6y
(3 + x)2
2
3
−1
2
0
5
1
1
−36
2
3
2
3
= 2π (3) − (3) − 6(3)
= π 4x − 2x + x −
2
3
3
3 + x −1
1
5
−36
1
2
3
−2π (2) − (2) − 6(2)
= π 4 · 0 − 2(0)2 + (0)3 −
2
3
3
3 + (0)
h
π
=
.
−π 4 · (−1) − 2(−1)2 +
3
1
−36 i
+ (−1)3 −
3
3 + (−1)
π
.
=
3
3
1
over the interval
ax2 + 1
0 ≤ x ≤ 1, where a > 0 is a constant. Find the volume of the solid formed by rotating the
bounded region about the y-axis. (Note: Your answer will possibly contain the constant a)
6. (Test Question) Consider the region under the graph y =
Solution: The axis of rotation is the y-axis and our variable of integration is x thus we
should use the shell methoda . Thus our volume is found by the integral
Z 1
1
2π
x 2
dx
ax + 1
0
Let u = ax2 + 1. Then du = 2ax dx. At x = 0, u = 1 and at x = 1, u = a + 1. Thus
Z
Z 1
π a+1 1
1
dx =
du
x 2
2π
ax + 1
a 1
u
0
a+1
π
=
ln |u|
a
1
π
π
=
ln |a + 1| − ln |1|
a
a
π
ln |a + 1|.
=
a
a
If we wanted to use disk method we would need to integrate over y.
4