Integrating surface integrals in polar coordinates.

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. :))

Solution 1:

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$