7.5
September 1, 2011
2
by symmetry, suppose it is (0,c)
9
Z
p
(y − c)2 9 − ydy = 0
0
let y = 9 sin2 θ
π/2
Z
Z
2
π/2
(9 sin θ)(3 cos θ)(18 sin θ cos θ)dθ = c
(3 cos θ)(18 sin θ cos θ)dθ
0
0
c = 2/5
5
by symmetry, suppose it is (0,c)
1
Z
p
(y − c)2 4 − (y + 1)2 dy = 0
0
let y + 1 = 2 sin θ
Z
π/2
(2 sin θ − 1 − c) cos2 θdθ = 0
π/6
√
c=
3
√ −1
4π/3 − 3)
8
the region consists of a semicircle and a rectangle, suppose the centroids
are (1/2, c1 ) and (c2 , −c2 ) respectively(pappus)
1
Z
0
p
4
(y − c1 )2 1 − y 2 dy = 0, c1 =
3π
Z
2
(x − c2 )(2 − x)dx = 0, c2 = 2/3
0
1
global centroid is
(1/2, 4/(3π))π/2 + (2/3, −2/3)2 = (π/4 + 4/3, −2/3)
10
2πcπr2 /4 = 2πr3 /3, c = 4r/3π
13
by symmetry, suppose it is (π/2, c)
Z 1
(y − c)(π − 2 sin−1 y)dy = 0
0
let y = sin x
π/2
Z
(sin x − c)(π − 2x) cos xdx = 0, c = π/8
0
16
by symmetry, suppose it is (0,c)
Z 2
Z 1
(y − c)2dy +
(y − c)2ydy = 0, c = 11/9
1
0
19
split into semicircle and rectangle, suppose centroids are c1 and −c2
respectively, by pappus
(2/3)π = (π/4)2πc1 , c1 = 4/3π
π/3 = 2πc2 (1/2)c2 = 1/3
the global centroid is
(4/3π)(2/3)π − (1/3)π/3 = 8/9 − π/9
22
the centroid of the line is (1/2, 1/2) the area is by pappus
p
√
√
22π 5/2 = 2 5π
25
suppose (c1 , c2 )
Z
Z
π/2
√
(x − c1 ) x cos xdx = 0
0
√
π/2 Z x cos x
y − c2 dydx = 0
0
28
Z
∞
2πe−x
2
0
q
2
1 + (2xe−x2 )2 dx < ∞ ∵ e−x x−2
−∞
the length is infinity hence no centroid
2