Calculus- Early Transcendentals, 2021a

500 Multiple Integration
Finally,
M y =
∫ π/2 ∫ cosx
−π/2
0
xdydx=
∫ π/2
−π/2
xcosxdx= 0.
So ¯x = 0 as expected, and ȳ = π/4/2 = π/8. This is the same problem as in Example 8.21; itmaybe
helpful to compare the two solutions.
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Example 14.8: Center of Mass of 2-D Plate
Find the center of mass of a two-dimensional plate that occupies the quarter circle x 2 + y 2 ≤ 1 in
the first quadrant and has density k(x 2 + y 2 ).
Solution. It seems clear that because of the symmetry of both the region and the density function (both
are important!), ¯x = ȳ. We’ll do both to check our work.
Jumping right in:
M =
∫ 1 ∫ √ 1−x 2
0
0
∫ 1
k(x 2 + y 2 )dydx = k x 2√ 1 − x 2 + (1 − x2 ) 3/2
dx.
0
3
This integral is something we can do, but it’s a bit unpleasant. Since everything in sight is related to a
circle, let’s back up and try polar coordinates. Then x 2 + y 2 = r 2 and
M =
∫ π/2 ∫ 1
0
0
∫ π/2
k(r 2 )rdrdθ = k
0
r 4
4
∣
1
0
∫ π/2 1
dθ = k
0 4 dθ = kπ 8 .
Much better. Next, since y = r sinθ,
∫ π/2 ∫ 1
∫ π/2
M x = k r 4 1
sinθ drdθ = k
(−
0 0
0 5 sinθ dθ = k 1 )
π/2
cosθ
5 ∣ = k
0
5 .
Similarly,
∫ π/2 ∫ 1
∫ π/2
M y = k r 4 1
cosθ drdθ = k
0 0
0 5 cosθ dθ = k 1 ∣ ∣∣∣
π/2
5 sinθ = k
0
5 .
Finally, ¯x = ȳ = 8
5π .
Exercises for 14.3
♣
Exercise 14.3.1 Find the center of mass of a two-dimensional plate that occupies the square [0,1] × [0,1]
and has density function xy.
Exercise 14.3.2 Find the center of mass of a two-dimensional plate that occupies the triangle 0 ≤ x ≤ 1,
0 ≤ y ≤ x, and has density function xy.
Exercise 14.3.3 Find the center of mass of a two-dimensional plate that occupies the upper unit semicircle
centered at (0,0) and has density function y.