SME
Calculus II · Problem type
Volume by washers
Use this for “The region between y = 3x and y = x² is turned about the x-axis. Find the volume of the solid.”
Diagram
The region between y = 2x and y = x² on [0, 2], turned about the x-axis.
At x = 1 the slice is a washer: R = 2, r = 1, area π(R² − r²) = 9.425.
V = ∫ π(f² − g²) dx from 0 to 2 = 13.4.
This is an example. Type your own values to start a new problem.
∫
0
π(
(
x)
2
−
(x²)
2
)
dx
=
V
x
Slice at
example
R
Outer radius (the line)
worked out (not typed)
r
Inner radius (the parabola)
worked out (not typed)
A
Area of the washer
worked out (not typed)
Formulas
R
=
k
×
x
2 = 2 × 1
r
=
x
2
1
=
1
2
A
=
π
×
(
R
2
−
r
2
)
9.4248
=
π
×
(
2
2
−
1
2
)
V
=
2
×
π
15
×
k
5
13.4041
=
2
×
π
15
×
2
5
Assumptions
•
The line y = kx and the parabola y = x² meet at x = 0 and x = k, and the line is on top between them. The slope k is more than 0.
•
A slice at x, dx thick, is a washer: a disk of outer radius R = kx with a hole of inner radius r = x², so its area is π(R² − r²).
•
Adding the washers from 0 to k gives V = ∫ from 0 to k of π((kx)² − (x²)²) dx. Square each radius first, then subtract.
Step-by-step
WE KNOW
k = 2, x = 1
FIND
outer radius (the line) (R), inner radius (the parabola) (r), area of the washer (A), volume of the solid (V)
Step 1 · Find outer radius (the line): R
Use R = k × x
The
line
is
farther
from
the
axis,
so
it
makes
the
outer
edge
of
the
washer:
R
=
kx.
R
=
2
×
1
R
=
2
Step 2 · Find inner radius (the parabola): r
Use r = x²
The
parabola
is
nearer
the
axis,
so
it
makes
the
edge
of
the
hole:
r
=
x
2
.
r
=
1
2
r
=
1
Step 3 · Find area of the washer (A)
Use A = π × (R² − r²)
The washer is the big disk with the small disk taken out. Square each radius before you subtract, not after.
A
=
π
×
(
2
2
−
1
2
)
A
=
(4
−
1)
×
π
A
=
3
×
π
A
=
9.4248
Step 4 · Find volume of the solid (V)
Use V = 2 × π ÷ 15 × k⁵
Integrate
the
washer
area
π(
k
2
x
2
−
x
4
)
from
0
to
k
.
At
k
it
is
π(
k
5
3
−
k
5
5
)
=
2π
k
5
15,
and
at
0
it
is
0.
V
=
2
×
π
15
×
2
5
∫
from
0
to
2
of
π
×
(
4
x
²
−
x
4
)
dx
=
[π
×
(
4
x
³
3
−
x
5
5
)]
from
0
to
2
=
13.4041
V
=
13.4041
Check
Put the numbers back in. Both sides should match.
2 = 2 × 1 ✓
1
=
1
2
✓
9.4248
=
π
×
(
2
2
−
1
2
)
✓
13.4041
=
2
×
π
15
×
2
5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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