# Surface Area using the integral

• Mar 6th 2010, 10:57 AM
mlckb1
Surface Area using the integral
let f(x)=sinx fpr 0 to pi. Find the surface Area of the solid formed by revolving about the x-axis.

Now i know that it is the integral of 2pi Y ds which becomes 2pi sinx(1+(sinx)^2)^(1/2)

i just don't know how to integrate it after that. i tried a u sub and that didn't work. if someone could point me in the right direction that'd be great.
• Mar 6th 2010, 11:05 AM
Krizalid
you made a mistake, the integral is actually $\displaystyle \int_0^\pi\sin(x)\sqrt{1+\cos^2x}\,dx,$ put $\displaystyle t=\cos x$ and the remaining integral can be tackled by using trig. or hyperbolic substitution.
• Mar 6th 2010, 11:08 AM
mlckb1
Quote:

Originally Posted by Krizalid
you made a mistake, the integral is actually $\displaystyle \int_0^\pi\sin(x)\sqrt{1+\cos^2x}\,dx,$ put $\displaystyle t=\cos x$ and the remaining integral can be tackled by using trig. or hyperbolic substitution.

Okay that was a really stupid mistake but that makes it waaaayyyy easier to solve. thanks