Download Calculus II, Section 6.3, #39 Volumes by Cylindrical Shells The

Calculus II, Section 6.3, #39
Volumes by Cylindrical Shells
The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting
solid by any method.1
y 2 − x2 = 1,
y = 2;
about the x-axis.
The curve y 2 − x2 = 1 is a hyperbola that opens in the y-direction. In Figure 1, the hyperbola and the line
y = 2 are sketched along with the axis of revolution. From the figure, the lower branch of the hyperbola is
not a boundary of the region. We solve the hyperbola equation for y:
y
2
y=2
2
2
y −x =1
y 2 = 1 + x2
y=
√
1 + x2
1
so
y=−
p
1 + x2
or y =
p
1 + x2
The region bounded by the curves is in Quadrants I
and II, so the values of y are all positive. Thus,
p
y = 1 + x2
Now we can find the points of intersection by solving
p
1 + x2 = 2
√
− 3
√
x
3
-1
√
y = − 1 + x2
-2
Figure 1: The curves and the region.
1 + x2 = 4
x2 = 3
so
√
√
x = − 3 or x = 3
√ √ Thus the points of intersection are − 3,2 and
3,2 .
For this problem, we are not told which of washers or shells we should use. (As we would encounter in any real
world problem.) Let’s set up the integral for both methods, and then we’ll decide which to compute.
In Figure 2, we have drawn a representative rectangle perpendicular to the axis of revolution. This will
generate√a washer of height
√ ∆x. The outside radius of the washer is rout = 2 − 0 = 2, while the inside radius
is rin = 1 + x2 − 0 = 1 + x2 . We have
p
2 2
2
Vwasher = π (2) − π
∆x
1+x
= 4π − π 1 + x2 ∆x
√
√
and we generate washers from x = − 3 to x = 3, so the volume of the solid of revolution, V , is given
by
V =
Z
√ 4π
x=− 3
Z x=√3
=π
1 Stewart,
√
x= 3
√
x=− 3
− π 1 + x2 dx
3 − x2 dx
Calculus, Early Transcendentals, p. 455, #39.
Calculus II
Volumes by Cylindrical Shells
y
y
y=2
y=2
2
2
y=
rout
√
1 + x2
x=−
1
p
y2 − 1
1
x=
p
y2 − 1
rin
rshell
√
x
√
3
− 3
√
√
− 3
-1
-1
-2
-2
Figure 2: The region and a representative
rectangle for a washer.
x
3
Figure 3: The region and a representative
rectangle for a shell.
In Figure 3, we’ve drawn a representive rectangle parallel to the axis of revolution. This will generate a shell
of thickness ∆y. The volume of the shell is given by
p
p
Vshell = (2πy)
y 2 − 1 − − y 2 − 1 (∆y)
We create shells from y = 1 to y = 2, so the volume, V , of the solid of revolution is given by
Z
V =
y=2
(2πy)
y=1
= 2π
Z
= 4π
Z
y=2
y=1
y=2
y·2
y
y=1
p
p
y2 − 1 − − y2 − 1
dy
p
y 2 − 1 dy
p
y 2 − 1 dy
So now we have the two integrals
V =π
Z
√
x= 3
2
√
x=− 3
3 − x dx
or V = 4π
Z
y=2
y=1
y
p
y 2 − 1 dy
The left (washer) integral is a basic antiderivative and fundamental theorem of calculus, while the right
(shell) integral requires a substitution to complete.
Let’s compute
V =π
Z
√
x= 3
√
x=− 3
3 − x2 dx
x=√3
x3
= π 3x −
3 x=−√3
"
√ 3 !#
√ 3 !
√
√
3
− 3
− 3·− 3−
=π
3 3−
3
3
"
√
√ #
√
√
3 3
3 3
=π 3 3−
+3 3−
3
3
h √ i
=π 4 3
√
= 4π 3
Calculus II
Volumes by Cylindrical Shells
Now let’s compute
Z y=2 p
y y 2 − 1 dy
V = 4π
y=1
Since we don’t know a basic antiderivative for the integrand y
so du = 2y dy.
1
2
V = 4π ·
= 2π
Z
Z
y=2
y=1
y=2 p
2·y
p
y 2 − 1, we’ll try substitution. Let u = y 2 − 1,
p
y 2 − 1 dy
y 2 − 1 · 2y dy
y=1
2
2
When y = 1, u = (1) − 1 = 0, and when y = 2, u = (2) − 1 = 3. We have
V = 2π
Z
u=3
√
u du
u=0
u=3
= 2π
Z
u1/2 du
u=0
u=3
u3/2
= 2π
3/2 u=0
3/2 u=3
2u
= 2π
3
u=0
2 · 33/2
2 · 03/2
= 2π
−
3
3
i
h √
= 2π 2 3 − 0
√
= 4π 3
Thus, the volume of the solid √
formed when the region bounded by the curves y 2 − x2 = 1 and y = 2 is
revolved about the x-axis is 4π 3 units3 .
Students often ask “So which method is better?” or “Which method is easier?” The answer to these
questions is “Neither.” For this problem, both methods gave us integrals that we were able to compute and
neither of the integrals was significantly more challenging.