18.01 PROBLEM SET 7 – SOLUTIONS

2. Cross-sectional area = (2x)^{2} = 4x^{2} = 4(1 − y^{/}2) = 4 − 2y^{2}.

V = ∫_{−\sqrt{2}}^{√\overline{2}}(4 − 2y^{2})dy = 2 ∫_{0}^{√\overline{2}}(4 − 2y^{2})dy = (8y − \frac{4}{3}y^{3}) |_{0}^{√\overline{2}} = \frac{16\sqrt{2}}{3}.

(5 points)

3. Since

\frac{df}{dx} = −2 · (−\frac{1}{2}) · x^{−3/2} = x^{−3/2},

the tangent line at x = 1 has the form y = x + b, so 1 + b = f(1) = −2.

Therefore the tangent line is y = x − 3.

\frac{d^{2}f}{dx^{2}} = −\frac{3}{2}x^{−5/2} < 0,

so f is concave down, and then its tangent at any point is above the graph of f(x), hence

x − 3 ≥ −2x^{−3/2} for all x > 0

with equality only for x = 1.

(5 points)

4.

(a) By the shell method:

V = 2 · ∫_{r}^{a} 2πx\sqrt{a^{2} − x^{2}}dx = −\frac{4π}{3} (a^{2} − x^{2})^{3/2}|_{r}^{a} = \frac{4π}{3} (a^{2} − x^{2})^{3/2}.

Surface area revolved around the y-axis:

A = 2 · ∫_{r}^{a} 2πx\sqrt{1 + (\frac{dy}{dx})^{2}}dx = 4πa · ∫_{r}^{a} \frac{xdx}{\sqrt{a^{2} − x^{2}}} =
= 4πa · (−\sqrt{a^{2} − x^{2}}) |_{r}^{a} = 4πa\sqrt{a^{2} − x^{2}}.

(b)

48π^{2} · V = 48π^{2} · \frac{4π}{3} (a^{2} − r^{2})^{3/2} = (4π)^{3}(a^{2} − r^{2})^{3/2} = (\frac{A}{a})^{3} .

Typeset by AMS-TEX 2 18.01 PROBLEM SET 7 – SOLUTIONS

(c)

\frac{dV}{dr} = \frac{4π}{3} · \frac{3}{2}(a^{2} − r^{2})^{1/2} · (−2r) = −4πr(a^{2} − r^{2})^{1/2}.
\frac{dA}{dr} = 4πa · \frac{1}{2}(a^{2} − r^{2})^{−1/2} · (−2r) = −4πra(a^{2} − r^{2})^{−1/2}.
\frac{dV}{dr} − \frac{dA}{dr} = −4πr(a^{2} − r^{2})^{−1/2}(a^{2} − r^{2} − a).

Hence \frac{d}{dr} (V − A) = 0 if r = 0, a, or, \sqrt{a^{2} − a}. Now

(V − A)|_{r=0} = \frac{4π}{3} a^{3} − 4πa^{2} = \frac{4πa^{2}(a − 3)}{3}
(V − A)|_{r=a} = 0
(V − A)|_{r=\sqrt{a^{2}−a}} = \frac{4π}{3/2}a^{3} − 4πa^{3/2} = −\frac{8π}{3} a^{3/2}.

By 3. a − 3 ≥ −2a^{−1/2}, so

\frac{4πa^{2}(a − 3)}{3} ≥ −\frac{8π}{3} a^{3/2}

with equality only for a = 1. This shows that the minimum of V − A is −\frac{8π}{3} a^{3/2} for r = \sqrt{a^{2} − a}.

(10 points)