Evaluate a surface integral: x^2 + z^2 = 1, y = 0, y = 1,

[hank]

\(\displaystyle \int\int_{\sigma} x^{2}y \ \ dS\)

\(\displaystyle \sigma\) is the portion of the cylinder \(\displaystyle x^{2} + z^{2} = 1\) between \(\displaystyle y = 0, y = 1\) and the xy-plane.

I get to this:
\(\displaystyle \int^{1}_{0} \int^{1}_{0} \ y\sqrt{1-z^{2}} \ dydz\)
but I'm not sure how to proceed from here.

Am I correct getting to this point?
If so, where do I go from here?

I experimented with converting to polar coordinates and ended up with this:
\(\displaystyle \int^{\pi}_{0} \int^{1}_{0} \ r^{3}\sin\theta \ drd\theta\)
but I'm pretty sure that's wrong as I end up with an answer of zero where the book tells me it should be \(\displaystyle \frac{\pi}{4}\).

I got all the other problems right, but can't get this one. Thanks in advance.

 

[galactus]

Re: Evaluate a surface integral

This is a surface integral. You are apparently not using the formula for surface integrals. Sorry, no I do not think you are correct to this point.

\(\displaystyle z=\sqrt{1-x^{2}}\), R is the rectangular region enclosed by x = -1 and x = 1, and y = 0.

\(\displaystyle \frac{{\partial}z}{{\partial}x}=\frac{-x}{\sqrt{1-x^{2}}}\) which does not exist along the boundaries \(\displaystyle \pm{1}\).

We avoid them by using \(\displaystyle x=\pm{x_{0}}\) as boundaries where \(\displaystyle x_{0}\) is slightly smaller than 1 and let \(\displaystyle x_{0}\) approach 1.

By the symmetry of the surface and the integrand we can integrate over the region R' enclosed by \(\displaystyle x=0, \;\ x=x_{0}, \;\ y=0, \;\ y=1\) and the double the result.

\(\displaystyle \int\int{x^{2}y}dS\)

\(\displaystyle =\lim_{x_{0}\to{1^{-}}}2\int\int{x^{2}y\sqrt{\frac{x^{2}}{1-x^{2}}+1}}dA\)

\(\displaystyle =\lim_{x_{0}\to{1^{-}}}2\int_{0}^{x_{0}}\int_{0}^{1}\frac{x^{2}y}{\sqrt{1-x^{2}}}dydx\)

\(\displaystyle =\lim_{x_{0}\to{1^{-}}}\int_{0}^{x_{0}}}\frac{x^{2}}{\sqrt{1-x^{2}}}dx\), use \(\displaystyle x=sin{\theta}\) to get:

\(\displaystyle \int\int{x^{2}y}dS=\lim_{{\theta}_{0}\to\frac{\pi}{2}^{-}}\int_{0}^{{\theta}_{0}}sin^{2}{\theta}d{\theta}\), \(\displaystyle {\theta}_{0}=sin^{-1}x_{0}\)

\(\displaystyle =\lim_{{\theta}_{0}\to\frac{\pi}{2}^{-}}\left(\frac{1}{2}{\theta}_{0}-\frac{1}{4}sin2{\theta}_{0}\right)=\frac{\pi}{4}\)