a81f9c…: “Beside a beautiful building”

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Compute the flux of the field $\vec {F}= \vector {-y,x,x}$ across the plane $\vec {P}(u,v) = \vector {u,v,3-2u+v}$ oriented in the direction of a vector with positive $z$ -component over the region $R = \{(u,v):-2\le u \le 1, 3\le v\le 4\}$ . $\text {Flux} = \answer {-21}$

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

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

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

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

So our integral is now: $\int _3^4 \int _{-2}^1 -2v\d u\d v$