VOLUME - HOW YOU SLICE IT
The volume of a solid can often be determined using calculus by slicing the solid into thin sections
whose geometry is simple enough that their volumes can be calculated easily. Summing up all
these volumes, i.e., integrating, then gives the volume of the whole solid.
Example 1
A manufacturer drills a hole of radius of 3 through the center of a metal sphere of radius 7.
Find the volume of the resulting bead (Fig. 1).
Solution:
y
3
7
x
Fig. 1
The problem can be reframed as a volume of revolution. We imagine the solid being
formed by revolving some region in the xy-plane, in this case the moon shaped region in
Fig. 2, about some line - the y-axis in our problem.
y
y
y
r=3
2
2
(3,  7 − 3 ) = (3, 2√10 )
R= 72 − y2
x2 + y2 = 72 or
x = 72 − y2
y + y
y
h = y
x
x
x
3
7
(3, − 72 − 32 ) = (3, −2√10 )
Fig. 2
Fig.3
Fig.4
Slicing the region in the xy-plane into thin rectangles and revolving each rectangle about the
y-axis, Figs. 2 & 3, yields a solid, Fig. 4, made up of a stack of thin disks with holes in their
centers, i.e., washers, that approximates the bead,. Letting the height of the washers
approach zero and adding up the volumes of all the washers gives the volume of the bead.
The volume of a washer with outer radius R and inner radius r is easily obtained from
the formula for the volume of a right circular cylinder, since the volume of a washer is
the difference between the volume of a right circular cylinder of height h and radius R and
the volume of a right circular cylinder of height h and radius r. That is, the volume of a
washer is given by V = πR2h − πr2h, or V = π{R2 − r2}h.
Revolving the thin horizontal rectangle shown in Fig. 2 about the y-axis produces a thin
washer (Fig. 3) whose inner radius is r = 3 ,
whose outer radius is R =  72 − y2 = √ 49 − y2
and
whose height is
h = y.
So this washer has a volume of
2 2
2
, = π [40 − y2] y
V = π{ [√49 − y ] - 3 }y = π{ [49 − y2] - 9}y
and adding up all
such washers' volumes, i.e., evaluating the integral  π [40 − y2] dy
between y = −210 and y = 2√10 , gives the volume of the bead,
320
 π 40 - y dy  3
2 10
V
2
10
- 2 10
Example 2
Set up an integral that would allow the calculation of the volume of the solid produced by
revolving the region in the xy-plane bounded by y = 0, y = √x, x = 0, and x = 4 about the
x-axis (Figs. 5 & 6).
y
y
y = √x
x
x
4
Fig.6
Fig.5
From Figs. 7 & 8, we see that R = √x , r = 0 (there's no hole in the disk), and h = x.
So the volume of a typical thin disk is V = π(√x )2x = π x x and the volume of the solid
of revolution is given by
y
y
4
V   π x dx
0
h = x
R = √x
y = √x
4
Fig.7
x
x
Fig.8
If a solid of revolution is produced by revolving some region in the xy-plane about some
line other than one of the coordinate axes, the volume may be found in a similar manner, but
care must be taken in going from the step in which the region is sliced into rectangles to the
step in which a disk or washer is formed by the revolution. Sketching the disk or washer is
essential in order to get the right expressions for the inner radius, if the disks have holes in
them, and the outer radius.
Example 3
Set up an integral that would allow the calculation of the volume of the solid produced by
revolving the region in the xy-plane bounded by y = 0, y = √x, x = 0, and x = 4 about the
line y = 5 (Figs. 9 & 10).
y
y
y=5
y=5
y = √x
x
4
x
Fig. 9
From Figs. 11 & 12,
and
Fig. 10
R = 5,
r = 5 − √x,
h = Δx,
ΔV = π(52 − [5 − √x] 2)Δx = π(10√x − x)Δx,
4


V   π 10 x  x dx .
so
0
y h = Δx
y
y=5
y=5
y = √x
4
Fig. 11
r = 5 − √x
x
Fig. 12
R=5
x
Of course, many solids are not solids of revolution, but their volumes are found using the same basic
technique (which dates back at least to the time of Archimedes). The solid is sliced into thin slices
having simple geometries, and then all the small volumes are added together to get the volume of the
whole solid.
Example 4
Set up an integral that would allow the calculation of the volume of the solid whose base lies
in the xy-plane and is bounded by the graphs of y = x + 6 and y = x2 − 6, and which has
cross sections taken perpendicular to the x-axis that are 30°−60° triangles with the
hypotenuse lying in the xy-plane and the vertex of the 30° angle lying above the vertex of the
60° angle in the xy-plane.
y
y
(4, 10)
0.5√3(12 + x − x2)
y=x+6
y = x2 − 6
(−3, 3)
12 + x − x2
x
x
0.5(12 + x − x2)
Fig. 13
Fig. 14
A typical triangular cross section is shown in Figs. 13 & 14 from two different views.
Fig. 13 shows the view from the top of the solid and Fig. 14 shows the view from the side,
with the x-axis pointing directly out of the paper. From the two figures, it is clear that the
hypotenuse has a length of
c = (x + 6) − (x2 − 6) = 12 + x − x2.
From our knowledge of the relationship of the sides of a 30°−60° triangle we can obtain the
lengths of the two legs of the triangle,
a = 0.5(12 + x − x2)
b = 0.5√3(12 + x − x2).
and
From this the volume of the triangular slice is determined to be
ΔV = (1/2)[ 0.5(12 + x − x2)][ 0.5√3(12 + x − x2)]Δx,
ΔV = 0.125√3(12 + x − x2)2Δx.
The volume is then given by the integral expression
4

V   0.125 3 12  x  x 2
-3

2
dx.
or