9 FURTHER APPLICATIONS OF INTEGRATION

CHAPTER 9
SAMPLE EXAM SOLUTIONS
3.
(c) Substitute u = cos x. Thendu =−sin xdx,so
S = − ∫ √
−1
1
2π 1 + u 2 du = 2π ∫ √
1
−1
1 + u 2 du
[ √
= 2π u 1 + u 2 + ln
y
1
0
y=Ïx
R
LÁ
y=x@
Lª
1 x
( √ )] 1
u 1 + u 2 = 2√ 2π + π ln
−1
P = L 1 + L 2 =
∫ 1
0
√
1 +
( 1
2 √ x
( √ )
√ 2+1
2−1
) 2 ∫ 1 √
dx + 1 + (2x) 2 dx
4. (a) The area enclosed by an equatorial circle is πR 2 . A sphere of radius R is formed by rotating the curve
y = √ R 2 − x 2 about the x-axis.
(b) The surface area is S =
(c)
∫ R
−R
∫ R
−R
∫ R
−R
2π √ ( )
R 2 − x
√1 2 −x 2
+ √ dx, so the claim means that
R 2 − x 2
2π √ ( )
R 2 − x
√1 2 −x 2
+ √ dx = 4πR 2 .
R 2 − x 2
2π √ ( )
R 2 − x
√1 2 −x 2
+ √ dx =
R 2 − x 2
∫ R
−R
2π √ R 2 − x 2 √
R 2
R 2 − x 2 dx
= ∫ R
−R2πRdx = 4πR2
5. (a) The arc length L = ∫ √
4
0
1 + [ f ′ (x) ] 2 dx,sosince2≤ f ′ (x) ≤ 4, we can say that 4 √ 5 ≤ L ≤
4 √ 17.
(b) Here L = ∫ 4
0
√1 + [ g ′ (x) ] √
2 dx ≥ 4, since 1 + [ g ′ (x) ] 2 ≥ 1forallx.
(c) The surface area S = ∫ 4
0
√1 2π f (x) + [ f ′ (x) ] 2 dx,sosince1≤ f (x) ≤ 3and−1 ≤ f ′ (x) ≤ 1
on the interval [0, 4], we can say 8π ≤ S ≤ 24 √ 2π.
6. (a) L = ∫ √
( )
1
2
0
1 + 13
x −2/3 dx
(b) The integral is improper because lim
interval of integration.
x→0 + √
1 +
(
13
x −2/3 ) 2
=∞. The integrand is unbounded on the
(c) Wecanseefromagraphof f (x) = 3√ x that the curve has finite length; therefore, the integral must
converge.
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