James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 8.3 Applications to Physics and Engineering 563
and so
x − M y b
xf sxd dx
y
m − a
−
y b
f sxd dx a
y b
xf sxd dx
a
y b
f sxd dx
a
y − M y b 1
x
m − a 2 f f sxdg2 dx
y b
f sxd dx
a
−
y b 1
a 2 f f sxdg2 dx
y b
f sxd dx
a
Notice the cancellation of the ’s. The location of the center of mass is independent of
the density.
In summary, the center of mass of the plate (or the centroid of 5) is located at the
point sx, yd, where
8
x − 1 A yb xf sxd dx y
a
− 1 1
A yb 2 f f a
sxdg2 dx
y
y= œ„„„„„ r@-≈
4r
”0, ’
3π
Example 4 Find the center of mass of a semicircular plate of radius r.
SOLUTION In order to use (8) we place the semicircle as in Figure 11 so that
f sxd − sr 2 2 x 2 and a − 2r, b − r. Here there is no need to use the formula to calculate
x because, by the symmetry principle, the center of mass must lie on the y-axis,
so x − 0. The area of the semicircle is A − 1 2 r 2 , so
_r
0
r
x
1
y − 1 A yr 2 f f 2r
sxdg2 dx
FIGURE 11
− 1
1
2 r ? 1 2 2 y r (sr 2 2 x 2 ) 2 dx
2r
− 2
r 2 y r
0 sr 2 2 x 2 d dx (since the integrand is even)
− 2
r 2Fr 2 x 2 x 3
r
3G0
− 2
r 2 2r 3
3 − 4r
3
The center of mass is located at the point s0, 4rys3dd.
n
Example 5 Find the centroid of the region bounded by the curves y − cos x, y − 0,
x − 0, and x − y2.
SOLUTION The area of the region is
A − y y2
0
cos x dx − sin xg 0
y2
− 1
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