I need to calculate the surface area obtained by rotating $\sin\pi x$, $0\le x \le 1$ about the x-axis. So the surface area equation i think i have to use is:
$$A = \int_0^1 2\pi y \sqrt{1+(dy/dx)^2} \, dx$$
so what I did so far is
$$A = \int_0^1 2\pi \sin{\pi x} \sqrt{1+(\pi \cos \pi x)^2} \, dx$$
and I let $\tan{\theta} = \pi \cos{\pi x}$ so
$$\begin{align*}
A &= 2\pi \int_0^1 \sqrt{1+\tan^2{\theta}} \frac{-sec^2{\theta}}{\pi^2} \, d\theta \\
&= -2\pi^3 \int_0^1 \sec^3{\theta} d\theta \\
&= -2\pi^3\left( \frac{1}{2}\sec{\theta}\tan{\theta} + \frac{1}{2} \ln|\sec\theta + \tan\theta| \right)
\end{align*}$$
From here I am kind of lost but I can tell it doesn't really compare to my prof's answer of
$$A = 2\sqrt{1+\pi^2} + \frac{2}{\pi} \ln{\left( \pi + \sqrt{1+\pi^2}\right)}$$
Best Answer
You're fine as far as $\tan{\theta} = \pi \cos{\pi x}$, so $\sec^2{\theta} \, d\theta = -\pi^2 \sin{\pi x}$. The limits require some more care: when $x=0$, $\tan{\theta} = \pi$, so we have the rather weird $\arctan{\pi}=P$, say, as the lower limit. $x=1$ gives $\theta=-P$ as the upper limit. Hence, switching the limits round, we have $$ A = 2\pi \int_{-P}^{P} \sqrt{1+\tan^2{\theta}} \frac{\sec^2{\theta}}{\pi^2} \, d\theta = \frac{2}{\pi} \int_{-P}^{P} \sec^3{\theta} \, d\theta $$ This is clearly even, so we can do it as twice the integral from $0$ to $P$ to save confusion.
You have the correct integral for $\sec^3$, so now it's a matter of sticking the limits into $$ 2\pi \left( \frac{1}{2}\sec{\theta} \tan{\theta} + \frac{1}{2} \log{(\sec{\theta}+\tan{\theta})} \right) $$ The bottom limit obviously gives $0$ since $\sec{0}=1,\tan{0}=0$. Now, $\sec{\theta} = \sqrt{1+\tan^2{\theta}}$ for $0\leqslant \theta \leqslant \pi/2$, so we have $$ \tan{P} = \pi, \quad \sec{P} = \sqrt{1+\pi^2}, \tag{1} $$ so we get the answer as $$ A = \frac{2}{\pi} \left( \sec{P} \tan{P} + \log{(\sec{P}+\tan{P})} \right), $$ and sticking the values in (1) into this gives you the result.