c59f04…: “Left of a wonderful ring”

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Compute the flux of the field $\vec {F}= \vector {0,0,z}$ across the outwardly oriented surface $\vec {P}(u,v) = \vector {u,v,1-u^2-v^2}$ over the region $R = \{(u,v):u^2+v^2 \le 1\}$ . $\text {Flux} = \answer {\pi /2}$

You need to set-up the integral: $\iint _R \vec {F} \dotp \uvec {n}\d S$

Remember that: $\uvec {n}\d S = \left (\pp [\vec {P}]{u}\cross \pp [\vec {P}]{v}\right )\d u\d v$

Computing parital derivatives: $\begin{align*} \pp[\vec{P}]{u} &= \vector{1,0,-2u}\\ \pp[\vec{P}]{v} &= \vector{0,1,-2v}\\ \end{align*}$

Computing the cross-product: $\left (\pp [\vec {P}]{u}\cross \pp [\vec {P}]{v}\right ) = \vector {2u,2v,1}$

The vector $\vector {2u,2v,1}$ is pointing according to the given orientation.

$\vec {F}(x,y,z) \dotp \vector {2u,2v,1} = z$

And $z = 1-u^2-v^2$ .

So our integral is now: $\iint _R \left (1-u^2-v^2\right )\d u\d v$

Since we are intgrating over a disk, use polar coordinates.

Our integral is now: $\int _0^{2\pi }\int _0^1 \left (1-r^2\right )r\d r\d \theta$