MATH 136
Volume of a Rotational Solid:
The Disc Method
Let R be the region between the graph of f (z) and the z -axis over an interval [ g , h ]. If
we rotate R around the z -axis, then we obtain a three-dimensional solid. The volume of
this solid is given by
h
Vol =
∫ π ( f (z))
2
dz units3.
g
This “disc” method applies in two cases for the x y plane. The region R could be
between the graph of f (x) and the x -axis, and then R is rotated around the x -axis. Or
R could be between the graph of f and the y -axis, and then R is rotated around the y axis. In this second case, we must make sure to write f as a function of y with limits of
integration along the y -axis.
€
€
f ( x)
b
R1
x
a
b
Rotate R1 around x -axis →
Vol = ∫ π ( f ( x))2 dx
a
Same graph written as x = g(y )
€
d
c
R2
g(y )
d
Vol = ∫ π (g( y)) 2 dy
Rotate R2 around y -axis →
Example. Consider f (x) =
c
x − 2 over the interval [6, 11].
(a) Let R1 be the region between the graph of f and the x -axis. Find the volume of the
solid obtained by rotating R1 around the x -axis.
(b) Let R2 be the region between the graph of f €and the y -axis. Find the volume of the
solid obtained by rotating R2 around
the y -axis.
€
Solution. (a) We can apply the disc method directly:
 x2
 11 65 π
Vol = ∫ π ( x − 2) dx = π ∫ (x − 2) dx = π 
=
units3.
− 2 x 
2
2

 6
6
6
11
2
11
Note: If we enter ( X − 2) into Y1, then we can integrate numerically with the
command fnInt(Y12, X, 6, 11) which yields 32.5. Then Vol = 32.5 π cubic units.
(b) Because R2 is between the graph of f and the y -axis and we rotate around the y axis, we still can apply the disc method. However we must re-write the function and
the limits in terms of y . Since y = x − 2 , we have y 2 = x − 2 and x = y 2 + 2 . The range
on the y -axis becomes [2, 3] (by evaluating y at x = 6 and at x = 11). So the volume of
the solid obtained by rotating R2 around the y -axis is
 y€5 4y 3

Vol = ∫ π (y + 2) dy = π ∫ (y + 4 y + 4) dy = π 
+
+ 4y
3
 5

2
2
3
2
3
2
2 €
4
3
=
2
1073 π
units3.
15
2
If we enter Y + 2 into Y1, then we can use the command fnInt(Y12, Y, 2, 3) which
gives 71.5333 . . . . = 1073/15. Then Vol = 1073 π / 15 cubic units.
Volumes of a Cone and of a Sphere
We can use the disc method to derive formulas for the volumes of a cone and of a
sphere. Consider first a cone of height h units with a maximum radius of r units.
r
(r, h)
h
€
y=
h
0
h
x for 0 ≤ x ≤ r
r
or
r
x = y for 0 ≤ y ≤ h
h
x - axis
r
h
x for 0 ≤ x ≤ r . The cone can be obtained by rotating
r
the region between the line and the y -axis around the y -axis. But before integrating,
r
we must write x as a function of y . So we use x = y for 0 ≤ y ≤ h . Then
h
h
h  r 2
2
πr
y3
1
Vol (Cone) = ∫ π  y  dy = 2 ×
= π r2h units 3 .
h
3 0 3
h
€
0
Now consider the line y =
For a sphere of radius r units, consider the quarter circle f ( x) = r2 − x 2 for
0 ≤ x ≤ r . We obtain a hemisphere by rotating the quarter circle around the x -axis. We
double its volume to obtain the volume of a sphere. Hence,
r
Vol (Sphere) = 2 × ∫ π
0
(
r
2
r2 − x 2 dx = 2π ∫ r 2 − x 2 dx€
)
0
(
)
3 r
3
 2
 3
x
 = 2π  2r  = 4π r units 3 .
= 2π  r x −
3  0
3

 3 