18.01 Solutions, Problem Set 7

• §7.5 #1 Find the length of the arc y^{2} = x^{3} between (0, 0) and (4, 8).

Solution: Apply the arclength formula

s = ∫_{a}^{b} \sqrt{1 + f′(x)^{2}} dx

with f(x) = x^{3/2}, a = 0, b = 4. Then f′(x) = \frac{3}{2}x^{1/2}, so s = ∫_{0}^{4} \sqrt{1 + \frac{9}{4}x} dx.

The substitution u = 1 + \frac{9}{4}x gives

∫ \sqrt{1 + \frac{9}{4}x} dx = ∫ \frac{4}{9}\sqrt{u} du = \frac{8}{27}u^{3/2} + C = \frac{8}{27}(1 + \frac{9}{4}x)^{3/2} + C,

so s = \frac{8}{27}(10^{3/2} − 1).

• §7.5 #2 Find the length of the arc y = \frac{1}{4}x^{4} + \frac{1}{8x^{2}} , 1 ≤ x ≤ 2.

Solution: Since \frac{dy}{dx} = x^{3} − \frac{1}{4}x^{−3}, and

1 + (x^{3} − \frac{1}{4}x^{−3})^{2} = x^{6} + \frac{1}{2} + \frac{1}{16}x^{−6} = (x^{3} + \frac{1}{4}x^{−3})^{2},

the arclength formula gives

s = ∫_{1}^{2}(x^{3} + \frac{1}{4}x^{−3}) dx = (\frac{1}{4}x^{4} − \frac{1}{8}x^{−2})]_{1}^{2} = 4 − \frac{1}{32} − \frac{1}{4} + \frac{1}{8} = \frac{123}{32}.

• §7.5 #10 The curve x^{2/3} + y^{2/3} = a^{2/3} is called an astroid or a hypercycloid of four cusps. Sketch it and find its total length.

Solution: It is simplest to find \frac{dy}{dx} by implicit differentiation, which gives

\frac{2}{3}x^{−1/3} + \frac{2}{3}y^{−1/3} \frac{dy}{dx} = 0,

so \frac{dy}{dx} = −y^{1/3}x^{−1/3}. We can then simplify

1 + (\frac{dy}{dx})^{2} = 1 + y^{2/3}x^{−2/3} = \frac{x^{2/3} + y^{2/3}}{x^{2/3}} = a^{2/3}x^{−2/3},

so the variable y doesn’t appear in the expression we need to integrate. The length of the arc in the first quadrant is then

s = ∫_{0}^{a} a^{1/3}x^{−1/3} dx = \frac{3}{2}a^{1/3}x^{2/3}]_{0}^{a} = \frac{3}{2}a,

so the total length is 6a.

• §7.5 #12a

Set up, but do not evaluate, the integral for the length of the arc y = \sqrt{x}, 1 ≤ x ≤ 4.

Solution: We have \frac{dy}{dx} = \frac{1}{2}x^{−1/2}, so

s = ∫_{1}^{4} \sqrt{1 + \frac{1}{4x}} dx.

• §7.6 #1

Find the area of the surface of revolution generated by revolving the arc

y = \frac{1}{4}x^{4} + \frac{1}{8x^{2}} , 1 ≤ x ≤ 2, about the y-axis.

Solution: Since we revolved about the y-axis, we have dA = 2πr ds where r = x and

ds = (x^{3} + \frac{1}{4}x^{−3}) dx

(from §7.5 #2). The surface area is therefore

∫_{1}^{2} 2πx(x^{3} + \frac{1}{4}x^{−3}) dx = ∫_{1}^{2} 2πx^{4} + \frac{1}{4}x^{−2}) dx
= 2π (\frac{1}{5}x^{5} − \frac{1}{4}x^{−1})]_{1}^{2}
= 2π(\frac{32}{5} − \frac{1}{8} − \frac{1}{5} + \frac{1}{4})
= \frac{253}{20}π.

• §7.6 #5 Find the area of the surface of revolution generated by revolving the arc y = x^{3}, 0 ≤ x ≤ 1, about the x-axis.

Solution: Since we revolved about the x-axis, we have dA = 2πr ds where r = y = x^{3} and

ds = \sqrt{1 + (3x^{2})^{2}} ds.

The surface area is therefore ∫_{0}^{1} 2πx^{3}\sqrt{1 + 9x^{4}} dx. Now make the substi-

tution u = 1 + 9x^{4}, du = 36x^{3} dx, so

∫ 2πx^{3}\sqrt{1 + 9x^{4}} dx = ∫ \frac{π}{18}u^{1/2} du = \frac{π}{27}u^{3/2} + C = \frac{π}{27}(1 + 9x^{4})^{3/2} + C.

This gives

A = \frac{π}{27}(1 + 9x^{4})^{3/2}]_{0}^{1} = \frac{π}{27}(10^{3/2} − 1)

for the surface area.

• §7.6 #8 The loop of 9y^{2} = x(3 − x)^{2} is revolved about the y-axis. Find the area of the surface generated in this way.

Solution: The loop is formed by the curve with 0 ≤ x ≤ 3. The top half of the loop has equation: y = \frac{1}{3}x^{1/2}(3 − x) = x^{1/2} − \frac{1}{3}x^{3/2}, so

1 + (\frac{dy}{dx})^{2} = 1 + (\frac{1}{2}x^{−1/2} − \frac{1}{2}x^{1/2})^{2}
= \frac{1}{4}x^{−1} + \frac{1}{2} + \frac{1}{4}x
= (\frac{1}{2}x^{−1/2} + \frac{1}{2}x^{1/2})^{2} .

This gives dA = 2πr ds = π(x^{1/2} +x^{3/2}) dx. The surface area is twice that of the top half of the surface, so

A = 2π ∫_{0}^{3}(x^{1/2} + x^{3/2}) dx

= 2π (\frac{2}{3}x^{3/2} + \frac{2}{5}x^{5/2}))]_{0}^{3}
= 2π(\frac{2}{3} · 3\sqrt{3} + \frac{2}{5} · 9\sqrt{3})
= \frac{56}{5}π\sqrt{3}.

• §7.6 #10

Consider a cylinder circumscribed about a sphere of radius a. Let two planes perpendicular to the axis intersect the sphere. If these planes are at a distance h apart, show that the area of the spherical zone that lies between them on the sphere is 2πah.

Solution: Aligning the axis with the x-axis, the surface is gotten by revolving an arc of the circle x^{2} + y^{2} = a^{2} about the x-axis. The arc runs

from, say, x = c to x = d, where d − c = h. Since \frac{dy}{dx} = −\frac{x}{y}, we have

1 + (\frac{dy}{dx})^{2} = 1 + \frac{x^{2}}{y^{2}} = 1 + \frac{x^{2}}{a^{2} − x^{2}} = \frac{a^{2}}{a^{2} − x^{2}} .

Therefore

dA = 2πy ds = 2π\sqrt{a^{2} − x^{2}} · \frac{a}{\sqrt{a^{2} − x^{2}}} dx = 2πa dx,

and A = ∫_{c}^{d} 2πa dx = 2πah.

• p. 585 #10

If a point moves along a polar curve r = f(θ) at a constant speed, and is also moving away from the origin at a constant speed, show that the curve must be the exponential spiral r = ae^{bθ}.

Solution: We are given that \frac{dr}{dt} and \frac{ds}{dt} = \sqrt{(dr/dt)^{2} + r^{2}(dθ/dt)^{2}} are

constants. So r \frac{dθ}{dt} must also be constant, and since \frac{dr}{dt} = \frac{dr}{dθ} \frac{dθ}{dt}, we see that r and \frac{dr}{dθ} are proportional. Solving the differential equation \frac{dr}{dθ} = br (where b is a constant) gives r = ae^{bθ} (where a is a constant).

• p. 585 #18 Find the length of the curve r = a sin^{3} \frac{1}{3}θ from θ = 0 to θ = 3π/2. What is the total length of this curve?

Solution: Since \frac{dr}{dθ} = a sin^{2}(\frac{1}{3}θ) cos(\frac{1}{3}θ), we have

ds = \sqrt{(\frac{dr}{dθ} )^{2} + r^{2}} dθ
= \sqrt{a^{2} sin^{4}(\frac{1}{3}θ) cos^{2}(\frac{1}{3}θ) + a^{2} sin^{6}(\frac{1}{3}θ)} dθ
= \sqrt{a^{2} sin^{4}(\frac{1}{3}θ)} dθ
= a sin^{2}(\frac{1}{3}θ) dθ.

The arclength is therefore ∫_{0}^{3π/2} a sin^{2}(\frac{1}{3}θ) dθ. (You can integrate this using the “half-angle” formula, but we haven’t discussed this yet, so you can leave it in this form.) The entire curve is given by running from θ = 0 to θ = 3π, so its length

is ∫_{0}^{3π} a sin^{2}(\frac{1}{3}θ) dθ.

• §16.4 #5 Find the area of the surface generated by revolving the cardioid r = a(1 − cos θ) about the x-axis. (Set up, but do not evaluate the integral.)

Solution: The surface is generated by revolving the arc which runs from θ = 0 to θ = π. Then dA = 2πy ds where

y = r sin θ = a sin θ(1 − cos θ)
and ds = \sqrt{r^{2} + (dr/dθ)^{2}} dθ
= \sqrt{a^{2}(1 − cos θ)^{2} + a^{2} sin^{2} θ} dθ
= a\sqrt{2 − 2 cos θ} dθ.

So the suface area is

A = ∫_{0}^{π} 2\sqrt{2}πa^{2} sin θ(1 − cos θ)^{3/2} dθ.

(We do know how to evaluate this one: set u = 1 − cos θ gives

∫ sin θ(1 − cos θ)^{3/2} dθ = \frac{2}{5}(1 − cos θ)^{5/2},

so A = 2\sqrt{2}πa^{2} \frac{2}{5} · 2^{5/2} = \frac{32}{5}πa^{2}.)

• §16.4 #11

Find the surface area generated by revolving the circle r = 2a cos θ about the line (a) θ = 0; (b) θ = π/2. (Set up, but do not evaluate the integrals.)

Solution:

In either case

ds = \sqrt{r^{2} + (dr/dθ)^{2}} dθ = \sqrt{4a^{2} cos^{2} θ + 4a^{2} sin^{2} θ} dθ = 2a dθ.

In (a), we use the arc from θ = 0 to θ = π/2, and dA = 2πy ds = 2π · 2a cos θ sin θ · 2a dθ, which gives

A = 8πa^{2} ∫_{0}^{π/2} cos θ sin θ dθ.

In (a), we use the arc from θ = 0 to θ = π, and dA = 2πx ds = 2π · 2a cos^{2} θ · 2a dθ, which gives

A = 8πa^{2} ∫_{0}^{π} cos^{2} θ dθ.

• §16.4 #12 Find the length of one turn of the spiral r = θ from θ = 0 to θ = 2π. (Set up, but do not evaluate the integral.)

Solution: We have ds = \sqrt{r^{2} + (dr/dθ)^{2}} dθ == \sqrt{θ^{2} + 1} dθ, so

s = ∫_{0}^{2π} \sqrt{θ^{2} + 1} dθ.

• §16.5 #1

Find the area enclosed by the cardioid r = a(1 + cos θ). (Set up, but do not evaluate the integral.)

Solution: Since dA = \frac{1}{2}r^{2} dθ, we have

A = ∫_{0}^{2π} \frac{1}{2}a^{2}(1 + cos θ)^{2} dθ.

• §16.5 #6 Find the area inside the smaller loop of the limacon r = 1 + 2 cos θ. (Set up, but do not evaluate the integral.)

Solution: The smaller loop is given by 2π/3 ≤ θ ≤ 4π/3 (where cos θ ≤ −\frac{1}{2}, so r ≤ 0). The area is therefore

A = ∫_{2π/3}^{4π/3} \frac{1}{2}(1 + 2 cos θ)^{2} dθ.

• §16.5 #9 Find the area that lies inside both curves r = a cos θ and r = a(1 − cos θ). (Set up, but do not evaluate the integral.)

Solution: The curves intersect at θ = 0 and θ = ±\frac{π}{3} . The top half of our region inside r = a(1 − cos θ) for 0 ≤ θ ≤ \frac{π}{3} , and inside r = a cos θ for \frac{π}{3} ≤ θ ≤ \frac{π}{2} . The area is therefore

A = ∫_{0}^{π/3} a^{2}(1 − cos θ)^{2} dθ + ∫_{π/3}^{π/2} a^{2} cos^{2} θ dθ.