# Math Help - area of surface of revolution

1. ## area of surface of revolution

Hi guys. I can't think of a way to start this integration - can someone set me on the proper path? Thanks!

What I have so far:

find the area of the surface obtained by rotating the curve around the x-axis

y=sin(pix), 0≤x≤1

dy/dx = picos(pix)

1 + (dy/dx)^2 = 1 + (pi^2)cos^2(pix)

integral 2(pi)sin(pix)√[1 + (pi^2)cos^2(pix)] = ?

2. Originally Posted by swicket
Hi guys. I can't think of a way to start this integration - can someone set me on the proper path? Thanks!

What I have so far:

find the area of the surface obtained by rotating the curve around the x-axis

y=sin(pix), 0≤x≤1

dy/dx = picos(pix)

1 + (dy/dx)^2 = 1 + (pi^2)cos(pix)

integral √[1 + (pi^2)cos(pix)] = ?
You forgot to square the cosine function.

-Dan

3. Originally Posted by topsquark
You forgot to square the cosine function.

-Dan
Whoops, that was an error in transcription. I've edited the post to include the missing information.

Thanks!

4. Originally Posted by swicket
Whoops, that was an error in transcription. I've edited the post to include the missing information.

Thanks!
If you are asking how to integrate

$\int\sin(x)\sqrt{1+\cos^2(x)}dx$

The answer is with relative difficulty. This one is really really ugly, from what i have seen. But you could always try trapezoid rule

The actual answer due to your specific definite integral is $\int_0^{1}2\pi\sin(\pi{x})\sqrt{1+\cos^2(\pi{x})}d x=\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)+2\sqrt{2}$

EDIT: sorry I did not see the coefficient on the cos², now I reevaluated it and it is too messy to type. but it is approximately 7.778

5. $

\int_0^{1}2\pi\sin(\pi{x})\sqrt{1+\cos^2(\pi{x})}d x=\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)+2\sqrt{2}
$

Hello, I appreciate the answer, but I was hoping someone could give me a hint on the mechanics. I'd like to solve this myself, but I've gone through several different processes and I'm still having problems. What sort of substitution am I looking at here?

Thanks,
swicket

6. Originally Posted by swicket and edited by Mr F
$

\int_0^{1}2\pi\sin(\pi{x})\sqrt{1+ \pi^2 \cos^2(\pi{x})} \, dx= .......
$

Hello, I appreciate the answer, but I was hoping someone could give me a hint on the mechanics. I'd like to solve this myself, but I've gone through several different processes and I'm still having problems. What sort of substitution am I looking at here?

Thanks,
swicket
First let $u = \pi \cos (\pi x)$.

Then $S = -\frac{2}{\pi} \int_{\pi}^{-\pi} \sqrt{1 + u^2} \, du = \frac{2}{\pi} \int_{-\pi}^{\pi} \sqrt{1 + u^2} \, du$ ......

7. Ah thank you! I was going down that road, but I must have made an error. Your guidance is much appreciated!