# Calculating a surface integral between a cone and a hemisphere

Let $S$ be the surface formed by the portions of the hemisphere $z=\sqrt{1-x^2-y^2}$ and of the cone $z =\sqrt{x^2+y^2}$ with $x^2+y^2\le\frac{1}{2}$. Calculate $\int_S\vec{F}\cdot d\vec{S}$ (with the normal outside) where

$$\vec{F}=(xz+e^{y\sin z},2yz+\cos xz,-z^2+e^x\cos y)$$

I'm using the divergence theorem, which would have $$\int_S\vec{F}\cdot d\vec{S}=\iiint_D div\vec{F}dV$$

Changing the original variables to spherical coordinates:

$$x=\frac{\sqrt{2}}{2}\sin\varphi\cos\theta,$$

$$y=\frac{\sqrt{2}}{2}\sin\varphi\sin\theta,$$

$$z=\frac{\sqrt{2}}{2}\cos\varphi.$$

And

$$div(\vec{F})=z+2z-2z=z.$$

Is it appropriate for you to use the divergence theorem?

How would the limits of integration in $\iiint_D div\vec{F}dV$?

• It would be quite a bit easier if you used cylindrical co-ordinates, then $$\rho \in [0, \frac 1{ \sqrt 2}] \\ \theta \in [0,2 \pi] \\ \rho < z < \sqrt{1-\rho^2}$$ – WW1 Jul 15 '18 at 2:49