Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
7.2
Volume of a Solid by the Method of Cross-Sections
In this section we study how to find the volume of a solid by the method of
cross-section or slicing. But first we remind the reader of the volume of a
right cylinder with base of area A and height h:
S = Ah.
Let R be a solid lying alongside some interval [a, b]. By a cross-section we
mean the intersection of the solid with a plane perpendicular to the solid.
Let A(x) denote the area of the cross section at x. (See Figure 7.2.1(a))
Partition the interval [a, b] into n equal subintervals each of length ∆x = b−a
n .
The mesh points can be found by means of the formula xi = a + i∆x, i =
0, 1, · · · , n. The planes that are perpendicular to the solid at the mesh points
divide the solid into n slices or slabs. See Figure 7.2.1(b). We choose n large
enough so that the slice between xi−1 and xi is so thin that it can be looked
at as a right cylinder of height ∆x and area of base equals A(x∗i ) at some
point x∗i in the interval [xi−1 , xi ]. See Figure 7.2.1(c). Thus, the volume of
the ith slice is
∆Vi = A(x∗i )∆x.
Adding the n slices, we find an approximation to the volume of the solid
given by
n
X
V ≈
A(x∗i )∆x.
i=1
A better approximation is obtained by letting n increases without bound.
Thus, we define the volume of the solid under consideration by the definite
integral
Z b
V =
A(x)dx.
a
If the solid lies alongside the y−axis, we use the formula
Z
V =
d
A(y)dy.
c
1
Figure 7.2.1
Example 7.2.1
Find the volume of a cone of radius r and height h.
Solution.
Place the cone with vertex at the origin and axis the x−axis. Consider a
cross section at x. See Figure 7.2.2.
Figure 7.2.2
Using similar triangles argument, the radius of the cross section at x is
r 2 x2
y = rx
h so that A(x) = π h2 . Thus,
Z
V =
h
π
0
r 2 x2
πr2 h
dx =
2
h
3
Example 7.2.2
Find the volume of a sphere with radius r.
2
Solution.
Place the sphere with center at the origin. Consider a cross section at x.
See Figure 7.2.3.
Figure 7.2.3
The area of the cross-section is
A(x) = πy 2 .
Using the√Pythagorean formula, the radius of the cross section (a circle) at
x is y = r2 − x2 so that A(x) = π(r2 − x2 ). Thus,
r
Z r
x3
4
2
2
2
V =
π(r − x )dx = π r x −
= πr3
3 −r
3
−r
Example 7.2.3
Find the volume of a pyramid with height h and square base having a side
of length s.
Solution.
Place the pyramid with vertex at the origin. The cross-section at x is a
square with side of length L. See Figure 7.2.4.
Figure 7.2.4
3
The area of this cross-section is
A(x) = L2 .
Using a similar triangles discussion, we find
L=
sx
.
h
Hence, the cross-section has the area
A(x) =
s2 x2
.
h2
The volume of the pyramid is
Z
V =
0
h
s2 x2
s2 x3
dx
=
h2
3h2
h
=
0
s2 h
3
Example 7.2.4
Find the volume of a right circular cylinder with height h and radius of base
r.
Solution.
Place the cylinder with one base at the origin as shown in Figure 7.2.5. The
cross-section at y is a circle with radius r so that A(y) = πr2 .
Figure 7.2.5
The volume of the cylinder is
Z
V =
h
πr2 dy = πr2 h
0
4
Volumes of Solids of Revolution
By a solid of revolution we mean a solid obtained by revolving a region
around a line. Consider the solid of revolution obtained by revolving a plane
region under the graph of f (x) around the x−axis. See Figure 7.2.6.
Each cross section is a circular disk of radius y, so its area is A(x) = πy 2 =
π[f (x)]2 . The volume of the solid is
b
Z
π[f (x)]2 dx.
V =
a
Figure 7.2.6
Example 7.2.5
√
The region bounded by the curve y = x + 1 and the x−axis between x = 0
and x = 9 is revolved around the x−axis. Find the volume of this solid of
revolution.
Solution.
The solid of revolution is given in Figure 7.2.7. A cross-section is a disk of
√
area A(x) = π( x + 1)2 . Thus, the total volume is given by
9
Z
V =
√
π( x + 1)2 dx =
0
x2 4 3
=π
+ x2 + x
2
3
9
Z
√
π(x + 2 x + 1)dx
0
9
=π
0
171
≈ 268.61 cubic units
2
Figure 7.2.7
5
If the revolution is performed around the y−axis, the roles of x and y are
interchanged so in that case the formula is
b
Z
πx2 dy,
V =
a
where x must be written as a function of y, i.e. x = f −1 (y).
Example 7.2.6
The curve y = x2 , 0 ≤ x ≤ 1 is rotated about the y-axis. Find the volume
of the resulting solid of revolution.
Solution.
The solid of revolution is shown in Figure 7.2.8.
Figure 7.2.8
A cross-section is a disk of area A(y) = πx2 = πy. Thus, the volume is
Z
V =
1
πydy = π
0
1
y 2 ≈ 1.571 cubic units
2 0
If the region being revolved is the area between two curves y = f (x) and
y = g(x), then each cross section is an annular ring (or washer) with outer
radius f (x) and inner radius g(x) (assuming f (x) ≥ g(x) ≥ 0.) See Figure
7.2.9.
Figure 7.2.9
6
The area of the annular ring is A(x) = π[(f (x))2 −(g(x))2 ], hence the volume
of the solid will be
Z b
Z b
π[f (x)2 − g(x)2 ]dx.
π[(ytop )2 − (ybottom )2 ]dx =
V =
a
a
If the revolution is performed around the y-axis, then
Z
V =
b
π[(xright )2 − (xlef t )2 ]dy.
a
Example 7.2.7
Find the volume of the solid obtained by revolving the area between y = x2
√
and y = x around the x-axis.
Solution.
First we need to find the intersection points of these curves in order to find
√
the interval of integration. Setting x2 = x and solving for x we find (0, 0)
and (1, 1). Hence, we must integrate from x = 0 to x = 1.
1
√
π[( x)2 − (x2 )2 ]dx =
0
1 2 1 5 1 3π
=
=π x − x
2
5
10
0
Z
Z
V =
7
0
1
π(x − x4 )dx