# Finding the Surface Area Obtained by Rotating $\sin(\frac{\pi}{3})x$ from $x=0$ to $x=3$

I tried using this formula

$$\int_a^bf(x)\sqrt{1+(f'(x))^2}dx$$

and managed to reduce the problem (after integrating by substitution twice) to: $$A=54(\pi)^2 \int_{46.32}^{-46.32} (\sec g)^3 \,dg$$ (where $u=\frac{3}{\pi}\tan g$, and $u=\cos (\frac{\pi}{3})x$. Now, to evaluate this I'll then have to use integration by parts which is, in this instant, quite unnecessarily tedious. So, I was wondering whether there is any quicker method to evaluate this. I have already thought of using Simpson's rule, but I don't believe that it is quite apt in this instant.

(Note:The answer Wolfram Alpha provided is $32.39$)

• I would consider the integral as a standard integral derived in an earlier chapter. Otherwise integration by part is the way to go with $f=secg$ and $g'=sec^2g$ – imranfat Mar 10 '16 at 18:13
• You can take half the integral and double it using a lower limit of zero, since the integrand is an even function. – John Molokach Mar 20 '16 at 22:02