1825e1…: “Through a rainy house”

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Compute the flux of the field $\vec {F}= \vector {x,y,z}$ across the outwardly oriented surface bounded above by $\vec {P}(u,v) = \vector {u,v,\sqrt {9-u^2-v^2}}$ and below by the plane $z=0$ . $\text {Flux} = \answer {54\pi }$

You will need to set-up two integrals, one for the top, and one for the bottom.

For the bottom, $\uvec {n} = \vector {0,0,-1}$ .

$\vec {F}\dotp \uvec {n}\d S = -z \d S$

However, since $z=0$ , no flux is contributed.

For the top, you could immediately switch to spherical coordinates.

The top is now parameterized by $\vec {P}(\theta ,\phi )=\vector {3\cos (\theta )\sin (\phi ),3\sin (\theta )\sin (\phi ),3\cos (\phi )}$ for $0\le \theta <2\pi$ and $0\le \phi \le \pi /2$ .

Computing, and dotting with $\vec {F}$ we find $\vec {F}\dotp \left (\pp [\vec {P}]{\phi }\cross \pp [\vec {P}]{\theta }\right ) = 27\sin (\phi )$

Now, $\iint _R \vec {F}\dotp \uvec {n}\d S = \int _0^{2\pi }\int _0^{\pi /2} 27\sin (\phi ) \d \phi \d \theta$