Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 7.3 TRIGONOMETRIC SUBSTITUTION ¤ 309
41. Let the equation of the large circle be x 2 + y 2 = R 2 . Then the equation of
the small circle is x 2 +(y − b) 2 = r 2 ,whereb = √ R 2 − r 2 is the distance
between the centers of the circles. The desired area is
A = r
√
−r b + r2 − x 2 − √ R 2 − x 2 dx
=2 r √
0 b + r2 − x 2 − √ R 2 − x 2 dx
=2 r
bdx+2 r
0 0
√
r2 − x 2 dx − 2 r
0
√
R2 − x 2 dx
The firstintegralisjust2br =2r √ R 2 − r 2 . The second integral represents the area of a quarter-circle of radius r,soitsvalue
is 1 4 πr2 . To evaluate the other integral, note that
Thus, the desired area is
√
a2 − x 2 dx = a 2 cos 2 θdθ [x = a sin θ, dx = a cos θdθ] = 1
2 a2 (1 + cos 2θ) dθ
= 1 2 a2 θ + 1 2 sin 2θ + C = 1 2 a2 (θ +sinθ cos θ)+C
= a2 x
2 arcsin + a2 x
√ a 2 − x 2
+ C = a2 x
a 2 a a
2 arcsin + x √
a2 − x
a 2 2 + C
A =2r √ R 2 − r 2 +2 1
4 πr2 − R 2 arcsin(x/R)+x √ R 2 − x 2 r
0
=2r √ R 2 − r 2 + 1 2 πr2 − R 2 arcsin(r/R)+r √ R 2 − r 2 = r √ R 2 − r 2 + π 2 r2 − R 2 arcsin(r/R)
43. We use cylindrical shells and assume that R>r. x 2 = r 2 − (y − R) 2 ⇒ x = ± r 2 − (y − R) 2 ,
so g(y) =2 r 2 − (y − R) 2 and
V = R+r
2πy · 2 r
R−r 2 − (y − R) 2 dy = r
4π(u + R)√ r
−r 2 − u 2 du [where u = y − R]
=4π r
u √ r
−r 2 − u 2 du +4πR
r
√
r2 − u
−r 2 where u = r sin θ , du = r cos θdθ
du
in the second integral
r
=4π − 1 3 (r2 − u 2 ) 3/2 +4πR π/2
−π/2 r2 cos 2 θdθ= − 4π (0 − 0) + 3 4πRr2 π/2
−π/2 cos2 θdθ
−r
=2πRr 2 π/2
−π/2 (1 + cos 2θ) dθ =2πRr2 θ + 1 2 sin 2θ π/2
−π/2 =2π2 Rr 2
Another method: Use washers instead of shells, so V =8πR r
r2 − y
0 2 dy as in Exercise 6.2.63(a), but evaluate the
integral using y = r sin θ.