# Find the surface integral

Find the surface integral of F= $(x,y,z)$ through the surface of $S = S_1 + S_2$ where $$S_1 \equiv z = 4 - x^2 - y^2, z \ge 0$$ and $S_2$ is the surface enclosed by $$x^2 + y^2 = 4$$

I have correctly found that $\int_{S_2} F dS = 0.$ However I am struggling to show that $\int_{S_1} F dS = 24\pi$. So far I have :

• Since $z \ge 0 \Rightarrow 4-x^2 - y^2 \ge 0$

• Parametrising gives $$\phi(u,v) = (ucos(v),usin(v),4-u^2-v^2)$$ where $u\in [0,2]$ and $v \in [0,2\pi]$

• $\frac{\partial \phi}{\partial u} = (cos(v),sin(v), -2u)$

• $\frac{\partial \phi}{\partial v}(-usin(v),-ucos(v), -2v)$

When finding the cross product I don't get something nice and hence I think I've gone wrong in parametrising (since i'm not great at that). Can someone explain why?

Your parameterization is wrong; you can check that $4-(\underbrace{u \cos v}_x)^2 - (\underbrace{u \sin v}_y)^2 \neq \underbrace{4-u^2-v^2}_{z}$.
I would use $\phi(x,y) = \langle x,y , 4-x^2-y^2 \rangle$, where $z \ge 0 \implies x^2 +y^2 \le 4$.