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Calculus II · Problem type

Volume by cylindrical shells

Use this for “The region under y = x(3 − x) is turned about the y-axis. Find the volume of the solid by shells.”

Diagram

yy = x(2 − x)2xr = x = 1h = 1
The region under y = x(2 − x) on [0, 2], turned about the y-axis.
The shell at x = 1 has radius 1 and height 1: unrolled, a sheet 2π × 1 × 1 = 6.283 by dx.
V = ∫ 2πx(f) dx from 0 to 2 = 8.378.
This is an example. Type your own values to start a new problem.
∫
0
2πx
·
x(
c
−
x)
dx
=
V
x
Shell at
example
h
Height of the shell
worked out (not typed)
S
Side area of the shell
worked out (not typed)

Formulas

h
=
x
×
(
c
−
x
)
1 = 1 × (2 − 1)
S
=
2
×
π
×
x
×
h
6.2832 = 2 × π × 1 × 1
V
=
π
6
×
c
4
8.3776
=
π
6
×
2
4

Assumptions

•
The arch y = x(c − x) meets the x-axis at x = 0 and x = c, and is above it between them. The width c is more than 0.
•
A strip at x, dx wide, turns about the y-axis into a shell of radius x and height f(x). Unrolled, it is a sheet 2πx long and f(x) tall.
•
Adding the shells from 0 to c gives V = ∫ from 0 to c of 2πx f(x) dx, so no curve needs solving for x.

Step-by-step

WE KNOW
c = 2, x = 1
FIND
height of the shell (h), side area of the shell (S), volume of the solid (V)
Step 1 · Find height of the shell (h)
Use h = x × (c − x)
The
shell
is
as
tall
as
the
arch
over
it,
so
its
height
is
the
curve’s
value
at
its
radius
x
.
h
=
1
×
(2
−
1)
h
=
1
Step 2 · Find side area of the shell (S)
Use S = 2 × π × x × h
Cut
the
shell
down
one
side
and
lay
it
flat:
a
sheet
as
long
as
the
circle
2πx
and
as
tall
as
h
.
S
=
2
×
π
×
1
×
1
S
=
6.2832
Step 3 · Find volume of the solid (V)
Use V = π ÷ 6 × c⁴
Integrate
the
side
area
2πx
·
x
(
c
−
x
)
=
2π(cx²
−
x
3
)
from
0
to
c
.
At
c
it
is
2π(
c
4
3
−
c
4
4
)
=
πc⁴
6,
and
at
0
it
is
0.
V
=
π
6
×
2
4
∫
from
0
to
2
of
2
×
π
×
(
2
x
²
−
x
3
)
dx
=
[2
×
π
×
(
2
x
³
3
−
x
4
4
)]
from
0
to
2
=
8.3776
V
=
8.3776
Check
Put the numbers back in. Both sides should match.
1 = 1 × (2 − 1) ✓
6.2832 = 2 × π × 1 × 1 ✓
8.3776
=
π
6
×
2
4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

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