Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 6.1 AREAS BETWEEN CURVES ¤ 269
49. By the symmetry of the problem, we consider only the first quadrant, where
y = x 2 ⇒ x = y. We are looking for a number b such that
b
0
ydy=
4
b
b 4
ydy ⇒
2
y 3/2 = 2 y 3/2 3
3
0 b
b 3/2 =4 3/2 − b 3/2 ⇒ 2b 3/2 =8 ⇒ b 3/2 =4 ⇒ b =4 2/3 ≈ 2.52.
⇒
51. We first assume that c>0, sincec can be replaced by −c in both equations without changing the graphs, and if c =0the
curves do not enclose a region. We see from the graph that the enclosed area A lies between x = −c and x = c, and by
symmetry, it is equal to four times the area in the first quadrant. The enclosed area is
A =4 c
0 (c2 − x 2 ) dx =4 c 2 x − 1 3 x3 c
0 =4 c 3 − 1 3 c3 =4 2
3 c3 = 8 3 c3
So A = 576 ⇔ 8 3 c3 =576 ⇔ c 3 =216 ⇔ c = 3√ 216 = 6.
Note that c = −6 is another solution, since the graphs are the same.
53. The curve and the line will determine a region when they intersect at two or
more points. So we solve the equation x/(x 2 +1)=mx
⇒
x = x(mx 2 + m) ⇒ x(mx 2 + m) − x =0 ⇒
x(mx 2 + m − 1) = 0 ⇒ x =0 or mx 2 + m − 1=0 ⇒
x =0or x 2 = 1 − m
m
1
⇒ x =0or x = ±
− 1. Notethatifm =1, this has only the solution x =0, and no region
m
is determined. But if 1/m − 1 > 0 ⇔ 1/m > 1 ⇔ 0