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 $z \ge 0 \implies x^2 +y^2 \le 4$.

  • $\begingroup$ Perfect, this works, thankyou! $\endgroup$ – user186609 Jan 15 '17 at 1:35

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