Solução_Calculo_Stewart_6e

F.
TX.10
PROBLEMS PLUS
1.
By symmetry, the problem can be reduced to finding the line x = c such that the shaded area is one-third of the area of the
quarter-circle.An equation of the semicircle is y = √ 49 − x 2 , so we require that c
√
0 49 − x2 dx = 1 · 1
3 4 π(7)2 ⇔
1 x √ 49 − x
2 2 + 49
2 sin−1 (x/7) c
= 49 π [by Formula 30] ⇔ 1 c √ 49 − c
0 12 2 2 + 49 2 sin−1 (c/7) = 49 π. 12
This equation would be difficult to solve exactly, so we plot the left-hand side as a function of c,andfind that the equation
holds for c ≈ 1.85. So the cuts should be made at distances of about 1.85 inches from the center of the pizza.
3. The given integral represents the difference of the shaded areas, which appears to
be 0. It can be calculated by integrating with respect to either x or y,sowefind x
in terms of y for each curve: y = 3√ 1 − x 7 ⇒ x = 7 1 − y 3 and
y = 7√ 1 − x 3 ⇒ x = 3 1 − y 7 ,so
1
3
0 1 − y7 − 7
1 − y 3 dy = 1√ 7
1 − x3 − 3√ 1 − x 7 dx. But this
equation is of the form z = −z. So 1√ 3
0 1 − x7 − 7√ 1 − x 3 dx =0.
0
5. The area A of the remaining part of the circle is given by
a
a2
A =4I =4 − x 2 − b
a2 − x
a 2 dx =4 1 − b a
a2 − x
a
2 dx
0
30
= 4 a (a − b) x
2
= 4 a (a − b)
0+ a2
2
√
a2 − x 2 + a2
2 sin−1 x
a
a
0
π
− 0
2
which is the area of an ellipse with semiaxes a and a − b.
= 4 a 2
a (a − b) π
= πa(a − b),
4
Alternate solution: Subtracting the area of the ellipse from the area of the circle gives us πa 2 − πab = πa (a − b),
as calculated above. (The formula for the area of an ellipse was derived in Example 2 in Section 7.3.)
0
351