Find the surface area of the portion of the hemisphere $f(x,y)=\sqrt{25-x^2-y^2}$ that lies above the cylinder $x^2+y^2=9$
To find my limits of integration I did the following:
$$x^2+y^2=9$$
$$y^2=9-x^2$$
$$y=\pm\sqrt{9-x^2}$$
And for $x$, this is $-3\le x\le3$
I then found the the partials as required by the surface area formula:
$$f_x=\frac{-x}{\sqrt{25-x^2-y^2}}$$
$$f_y=\frac{-y}{\sqrt{25-x^2-y^2}}$$
So I then formed the following double integral:
$$\int^3_{-3}\int^{\sqrt{9-x^2}}_{-\sqrt{9-x^2}}\sqrt{\frac{x^2}{25-x^2-y^2}+\frac{y^2}{25-x^2-y^2}+1}\;dy\;dx$$
Simplifying:
$$\int^3_{-3}\int^{\sqrt{9-x^2}}_{-\sqrt{9-x^2}}\sqrt{\frac{x^2+y^2}{25-x^2-y^2}+1}\;dy\;dx$$
Is this double integral set up properly? If so, hints on how I can solve this double integral? I tried converting to polar coordinates but was lost as to how to find the limits of integration.
I know the outer limit would be $\int^{2\pi}_{0}$ but the inner limit I'm lost.
Based on the following I'm taking a guess that the inner limit would be $\int^{3}_0$
$$r\;sin\theta=\sqrt{9-r^2cos^2\theta}$$
And when solving this I get $r=3$
Best Answer
Substituting $x=r\cos\theta$ and $y=r\sin\theta$, $$\int_0^{2\pi}\int_0^3\frac{5r}{\sqrt{5^2-r^2}}drd\theta=\int_0^{2\pi}\left[5\sqrt{5^2-r^2}\right]_3^0d\theta=10\pi$$ Using the already known formula for spherical cap (https://en.wikipedia.org/wiki/Spherical_cap) and $r=5,h=1$,
$$A=2\pi rh=10\pi$$ Those two results match.