surface integrals- spherical cap

[SigepBrandon]

Problem:

Consider the sphere x[sup:93998294]2[/sup:93998294]+y[sup:93998294]2[/sup:93998294]+z[sup:93998294]2[/sup:93998294]=4a[sup:93998294]2[/sup:93998294] of radius 2a and center at the origin. Use a surface integral to find a formula for the area of the cap, S, of the sphere above the disk x[sup:93998294]2[/sup:93998294]+y[sup:93998294]2[/sup:93998294]=a[sup:93998294]2[/sup:93998294] in the xy-plane. (Hint: use polar coordinates.)

after some examination I think the limits are r from a to 2a and theta from 0 to 2pi... but I'm not sure where to go from there or if that is even correct. most of the stuff I've seen on spherical caps has been in 3 space, so it's been difficult for me to visualize what's going on in 2 space (polar coordinates)

any guidance is greatly appreciated-
BS

 

[galactus]

We want the surface area of the portion of the sphere \(\displaystyle f(x,y)=x^{2}+y^{2}+z^{2}=4a^{2}\) above \(\displaystyle x^{2}+y^{2}=a^{2}\).

f(x,y) can be written as \(\displaystyle z=\sqrt{4a^{2}-x^{2}-y^{2}}\)

The first partial derivative:

\(\displaystyle f_{x}=\frac{-x}{\sqrt{4a^{2}-x^{2}-y^{2}}}\)

\(\displaystyle f_{y}=\frac{-y}{\sqrt{4a^{2}-x^{2}-y^{2}}}\)

The formula for surface area is \(\displaystyle dS=\int\int\sqrt{1+[f_{x}]^{2}+[f_{y}]^{2}}\)

Change to polar by letting \(\displaystyle x=rcos\theta, \;\ y=rsin\theta\).

This gives \(\displaystyle 2a\int_{0}^{2\pi}\int_{0}^{a}\frac{r}{\sqrt{4a^{2}-r^{2}}}drd\theta\)

This is what needs evaluated.