Early Transcendentals - 9.10.8

[Nick Rawlings]

This is an excellent exercise!  The surface area integral was straight forward but the subsequent algebra took me through a few wrong turns before I got it right.

I'm posting my solution here if anyone wants to suggest improvements.

\begin{equation*}

\begin{split}

y & = b\sqrt{1-\frac{x^2}{a^2}} = b \sqrt{\frac{a^2-x^2}{a^2}} = \frac{b}{a} \sqrt{a^2 - x^2} \\

\frac{dy}{dx} & = \frac{-bx}{a \sqrt{a^2 - x^2}} \\

SA & = \int_{-a}^{a} 2 \pi \frac{b}{a} \sqrt{a^2 - x^2} \sqrt{1 + \Big( \frac{-bx}{a \sqrt{a^2 - x^2}} \Big)^2} \, dx \\

& = \frac{2 \pi b}{a} \int_{-a}^{a} \sqrt{a^2 - x^2 + (a^2 - x^2) \frac{b^2 x^2}{a^2 (a^2 - x^2)}} \\

& = \frac{2 \pi b}{a} \int_{-a}^{a} \sqrt{\frac{a^4 - a^2x^2 + b^2 x^2}{a^2}} \\

& = \frac{2 \pi b}{a^2} \int_{-a}^{a} \sqrt{a^4 - a^2x^2 + b^2 x^2} \\

\end{split}

\end{equation*}

For $a > b$ we arrange the square root as $\sqrt{a^4 - (a^2 - b^2)x^2}$. 

\newline

For $a < b$ we arrange the square root as  $\sqrt{a^4 + (b^2 - a^2)x^2}$.

\newline

Continuing for $a > b$

\begin{equation*}

\begin{split}

SA & = \frac{2 \pi b}{a^2} \int_{-a}^{a} \sqrt{a^4 - (a^2 - b^2)x^2} \\

x & = \frac{a^2 \sin(u)}{\sqrt{a^2 - b^2}} \\

dx & = \frac{a^2 \cos(u)}{\sqrt{a^2 - b^2}} \, du \\

u & = \arcsin(\frac{\sqrt{a^2 - b^2}}{a^2} x) \\

SA & = \frac{2 \pi b}{a^2} \int_{-u}^{u} \sqrt{a^4 - (a^2 - b^2) \frac{a^4 \sin^2(u)}{a^2 - b^2}} \, \frac{a^2 \cos(u)}{\sqrt{a^2 - b^2}} \, du \\

SA & = \frac{2 \pi b}{a^2} \int_{-u}^{u} a^2 \cos(u) \, \frac{a^2 \cos(u)}{\sqrt{a^2 - b^2}} \, du \\

& = \frac{2 \pi a^2 b}{\sqrt{a^2 - b^2}} \int_{-u}^{u} \cos^2(u) \, du \\

& = \frac{2 \pi a^2 b}{\sqrt{a^2 - b^2}} \int_{-u}^{u} \frac{1}{2} (1 + \cos(2u)) \, du \\

& = \frac{\pi a^2 b}{\sqrt{a^2 - b^2}} \Big[ u + \frac{1}{2} \sin(2u) \Big]_{-u}^{u} \\

& = \frac{\pi a^2 b}{\sqrt{a^2 - b^2}} \Big[ \arcsin(\frac{\sqrt{a^2 - b^2}}{a^2} x) + \frac{1}{2} \sin(2 \arcsin(\frac{\sqrt{a^2 - b^2}}{a^2} x)) \Big]_{-a}^{a} \\

& = \frac{\pi a^2 b}{\sqrt{a^2 - b^2}} \Big[ \arcsin(\frac{\sqrt{a^2 - b^2}}{a^2} x) + \sin(\arcsin(\frac{\sqrt{a^2 - b^2}}{a^2} x)) \cos(\arcsin(\frac{\sqrt{a^2 - b^2}}{a^2} x)) \Big]_{-a}^{a} \\

& \sin(-a) = -\sin(a), \, \cos(-a) = \cos(a), \, {\rm by \, pythagoras} \, \cos(\arcsin(\frac{\sqrt{a^2 - b^2}}{a})) = \frac{b}{a} \\

SA & = \frac{\pi a^2 b}{\sqrt{a^2 - b^2}} \bigg( \arcsin(\frac{\sqrt{a^2 - b^2}}{a}) + \frac{\sqrt{a^2 - b^2}}{a} \frac{b}{a} + \arcsin(\frac{\sqrt{a^2 - b^2}}{a}) + \frac{\sqrt{a^2 - b^2}}{a} \frac{b}{a} \bigg) \\

& = \frac{2 \pi a^2 b}{\sqrt{a^2 - b^2}} \bigg( \arcsin(\frac{\sqrt{a^2 - b^2}}{a}) + \frac{b\sqrt{a^2 - b^2}}{a^2} \bigg) \\

& = \frac{2 \pi a^2 b}{\sqrt{a^2 - b^2}} \arcsin\bigg( \frac{\sqrt{a^2 - b^2}}{a} \bigg) + 2 \pi b^2 \\

\end{split}

\end{equation*}

Continuing for $a < b$

\begin{equation*}

\begin{split}

SA & = \frac{2 \pi b}{a^2} \int_{-a}^{a} \sqrt{a^4 + (b^2 - a^2)x^2} \\

x & = \frac{a^2 \tan(u)}{\sqrt{b^2 - a^2}} \\

dx & = \frac{a^2 \sec^2(u)}{\sqrt{b^2 - a^2}} \, du \\

u & = \arctan \bigg(\frac{\sqrt{b^2 - a^2}}{a^2} x \bigg) \\

SA & = \frac{2 \pi b}{a^2} \int_{-a}^{a} \sqrt{a^4 + (b^2 - a^2) \frac{a^4 \tan^2(u)}{b^2 - a^2}} \, \frac{a^2 \sec^2(u)}{\sqrt{b^2 - a^2}} \, du \\

SA & = \frac{2 \pi a^2 b}{\sqrt{b^2 - a^2}} \int_{-a}^{a} \sec^3(u) \, du \\

& {\rm wolframalpha.com} \int \sec^m(x) \, dx = \frac{\sin(x) \sec^{m-1}(x)}{m - 1} + \frac{m-2}{m-1} \int \sec^{-2 + m}(x) \, dx \, {\rm where } \, m = 3 \\

SA & = \frac{\pi a^2 b}{\sqrt{b^2 - a^2}} \bigg( \bigg[ \sin(u) \sec^2(u) \bigg] + \int \sec(u) \, du \bigg) \\

& = \frac{\pi a^2 b}{\sqrt{b^2 - a^2}} \bigg[ \tan(u) \sec(u) + \ln(\tan(u) + \sec(u)) \bigg] , \, {\rm substitute} \, u \\

& = \frac{\pi a^2 b}{\sqrt{b^2 - a^2}} \bigg[ \tan(\arctan \bigg(\frac{\sqrt{b^2 - a^2}}{a^2} x \bigg)) \sec(\arctan \bigg(\frac{\sqrt{b^2 - a^2}}{a^2} x \bigg)) + \\ 

& \ln(\tan(\arctan \bigg(\frac{\sqrt{b^2 - a^2}}{a^2} x \bigg)) + \sec(\arctan \bigg(\frac{\sqrt{b^2 - a^2}}{a^2} x \bigg))) \bigg]_{-a}^{a} \\

& \tan(-a) = -\tan(a), \, \sec(a) = \sec(-a), {\rm by \, pythagoras} \, \sec(\arctan(\frac{\sqrt{b^2 - a^2}}{a})) = \frac{b}{a} \\

SA & = \frac{\pi a^2 b}{\sqrt{b^2 - a^2}} \bigg( 

\frac{\sqrt{b^2 - a^2}}{a} \frac{b}{a} + \ln\Big( \frac{\sqrt{b^2 - a^2}}{a} + \frac{b}{a} \Big) +

\frac{\sqrt{b^2 - a^2}}{a} \frac{b}{a} + \ln\Big( \frac{\sqrt{b^2 - a^2}}{a} + \frac{b}{a} \Big) \bigg) \\

& = \frac{2 \pi a^2 b}{\sqrt{b^2 - a^2}} \bigg( \frac{b \sqrt{b^2 - a^2}}{a^2} + \ln\Big( \frac{\sqrt{b^2 - a^2}}{a} + \frac{b}{a} \Big) \bigg) \\

& = 2 \pi b^2 +  \frac{2 \pi a^2 b}{\sqrt{b^2 - a^2}} \ln\Big( \frac{\sqrt{b^2 - a^2}}{a} + \frac{b}{a} \Big) \\

\end{split}

\end{equation*}