# Integrating surface integrals in polar coordinates.

multivariable-calculuspolar coordinatessurface-integrals

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,

$$\displaystyle \int_0^{2\pi}\int_0^1 r ~ dr ~ d\theta = \pi$$