Solução_Calculo_Stewart_6e

F.
368 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION
TX.10
y = 1 A
2
0
= 1 A · 1
2
1
2 [(2x ) 2 − (x 2 ) 2 ] dx = 1 2
1
A
2 (22x − x 4 ) dx = 1
0
A · 1
2
16
2ln2 − 32
5 − 1
= 1 2ln2 A
Thus, the centroid is (x, y) ≈ (0.781, 1.330).
2
2x
2
2ln2 − x5
5
0
15
4ln2 − 16
≈ 1 (2.210106) ≈ 1.330
5 A
Since the position of a centroid is independent of density when the density is constant, we will assume for convenience that ρ =1in Exercises 38
and 39.
39. Choose x-andy-axes so that the base (one side of the triangle) lies along
the x-axis with the other vertex along the positive y-axisasshown. From
geometry, we know the medians intersect at a point 2 of the way from each
3
vertex (along the median) to the opposite side. The median from B goes to
the midpoint 1
2 (a + c), 0 of side AC, so the point of intersection of the
medians is 2
3 · 1
2 (a + c), 1 3 b = 1
3 (a + c), 1 3 b .
This can also be verified by finding the equations of two medians, and solving them simultaneously to find their point of
intersection. Now let us compute the location of the centroid of the triangle. The area is A = 1 (c − a)b.
2
x = 1 0
x · b
c
A a (a − x) dx +
a
0
x · b
c (c − x) dx
= 1 A b
a
=
Aa b 1
2 ax2 − 1 0
3 x3 + b 1
a
Ac
2 cx2 − 1 c
3 x3 = b − 1
0
Aa 2 a3 + 1
3 a3
=
0
a
(ax − x 2 ) dx + b c
2
a (c − a) · −a3
6 + 2
c (c − a) · c3 6 = 1
3(c − a) (c2 − a 2 )= a + c
3
+ b Ac
c
0
1
2 c3 − 1 3 c3
cx − x
2
dx
and
y = 1 A
0
a
2
1 b c
2
2 a (a − x) 1 b
dx + dx
0 2 c (c − x)
= 1 b
2
0
c
(a 2 − 2ax + x 2 ) dx + b2
(c 2 − 2cx + x 2 ) dx
A 2a 2 2c 2
b
2
= 1 A
= 1 b
2
A
a
2a 2 a 2 x − ax 2 + 1 3 x3 0
a + b2
2c 2 c 2 x − cx 2 + 1 3 x3 c
0
2a 2
−a 3 + a 3 − 1 3 a3 + b2
2c 2
c 3 − c 3 + 1 3 c3 = 1 A
0
b
2
6 (−a + c) =
2 (c − a)b2 · = b (c − a) b 6 3
a + c
Thus, the centroid is (x, y) =
3 , b
, as claimed.
3
Remarks: Actually the computation of y is all that is needed. By considering each side of the triangle in turn to be the base,
we see that the centroid is 1 3
of the way from each side to the opposite vertex and must therefore be the intersection of the
medians.