410 Chapter 6: Applications of Definite Integrals
Proof We draw the axis of revolution as the x-axis with the region Rin the first quadrant
(Figure 6.56). We let L(y) denote the length of the cross-section of Rperpendicular to the
y-axis at y. We assume L(y) to be continuous.
By the method of cylindrical shells, the volume of the solid generated by revolving
the region about the x-axis is
(10)
The y-coordinate of R’s centroid is
so that
Substituting for the last integral in Equation (10) gives With equal to
we have
EXAMPLE 6 Find the volume of the torus (doughnut) generated by revolving a circular
disk of radius aabout an axis in its plane at a distance from its center (Figure 6.57).
Solution We apply Pappus’s Theorem for volumes. The centroid of a disk is located at
its center, the area is is the distance from the centroid to the axis of
revolution (see Figure 6.57). Substituting these values into Equation (9), we find the
volume of the torus to be
The next example shows how we can use Equation (9) in Pappus’s Theorem to find one
of the coordinates of the centroid of a plane region of known area Awhen we also know the
volume Vof the solid generated by revolving the region about the other coordinate axis. That
is, if is the coordinate we want to find, we revolve the region around the x-axis so that
is the distance from the centroid to the axis of revolution. The idea is that the rotation
generates a solid of revolution whose volume Vis an already known quantity. Then we can
solve Equation (9) for which is the value of the centroid’s coordinate
EXAMPLE 7 Locate the centroid of a semicircular region of radius a.
Solution We consider the region between the semicircle (Figure 6.58) and
the x-axis and imagine revolving the region about the x-axis to generate a solid sphere. By
symmetry, the x-coordinate of the centroid is With in Equation (9), we have
y=V
2pA=s4>3dpa3
2ps1>2dpa2=4
3p a.
y=rx=0.
y=2a2-x2
y.r,
y=r
y
V=2psbdspa2d=2p2ba 2.
A=pa2, and r=b
bÚa
V=2prA.
y,rV=2pyA.Ay
Ld
c
y Lsyd dy =Ay.
y
'=y, dA =L(y) dyy=Ld
c
y
' dA
A=Ld
c
y Lsyd dy
A,
V=Ld
c
2psshell radiusdsshell heightd dy =2pLd
c
y Lsyd dy.
THEOREM 1 Pappus’s Theorem for Volumes
If a plane region is revolved once about a line in the plane that does not cut
through the region’s interior, then the volume of the solid it generates is equal to
the region’s area times the distance traveled by the region’s centroid during the
revolution. If is the distance from the axis of revolution to the centroid, then
(9)V=2prA.
r
x
y
d
y
c
0
L(y)
R
Centroid
⫽
FIGURE 6.56 The region Ris to be
revolved (once) about the x-axis to
generate a solid. A 1700-year-old theorem
says that the solid’s volume can be
calculated by multiplying the region’s area
by the distance traveled by its centroid
during the revolution.
Area: pa2
Circumference: 2pa
Distance from axis of
revolution to centroid
a
b
y
z
x
FIGURE 6.57 With Pappus’s first
theorem, we can find the volume of a torus
without having to integrate (Example 6).
Centroid
a
–aa0
a
x
y
3
4
FIGURE 6.58 With Pappus’s first
theorem, we can locate the centroid of a
semicircular region without having to
integrate (Example 7).
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