The question I have is to find the double integral of $z$ over $S$. Where $S$ is the hemispherical surface given by $x^2+y^2+z^2=1$ with $z \geq 0$.
I drew this out and found it to be the top half of a sphere and found the the cross product of tangent vectors to be $\frac{1}{z}$.
So my integral became $1$.
Using polar coordinates I found the circle projection of the sphere on the xy plane and got $0 \leq \theta \leq 2 \pi$ and $0 \leq r \leq 1$.
However when I integrate this I get $2 \pi$. Which I know is wrong as the circle with radius one has area $\pi$.
I can't figure out where my polar coordinates integral is wrong so any help would be greatly appreciated.
(apologies for the bad formatting, I'm relatively new and still figuring out the basics. :))
Best Answer
The surface is $z^2 = 1 - x^2 - y^2, z \geq 0$
$ \displaystyle \frac{\partial z}{\partial x} = \frac{x}{z}, \frac{\partial z}{\partial y} = \frac{y}{z}$
$ \displaystyle dS = \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} ~ dA= \frac{1}{z} ~ dA$
So the surface integral is,
$ \displaystyle \iint_S z ~dS = \iint_{x^2+y^2 \leq 1} 1 ~ dx ~ dy$
Converting it into polar coordinates,
$ \displaystyle \int_0^{2\pi}\int_0^1 r ~ dr ~ d\theta = \pi$