Here is the surface I'm trying to calculate the flux through:
I believe the area of the surface is $2\sqrt{20}$ (since the long side length is $\sqrt{2^2 + 4^2} =\sqrt{20}$)
$\vec F = \vec i + 2\vec j -3\vec k$. I want to calculate $\int_S \vec F \cdot d\vec A$, where $S$ is the oriented surface in the picture above.
What I've been doing so far is simplifying the dot product inside the integral based on the normal vector of $S$ (which is usually either positive or negative $\vec i, \vec j, $ or $\vec k$).
But in this case the normal vector to $S$ is not one of the three basis vectors, so I can't simplify the dot product inside the integral.
How can I proceed? Any help is appreciated.
Here the area vector $\vec A$ is a constant, as is $\vec F$, so the surface integral is just equal to $\vec F \cdot \vec A$.
Find $\vec A$ by finding a normal vector to the plane by taking the cross product of two non-parallel vectors in the plane (example $\begin{pmatrix} 0 \\\ -2 \\\ 4\end{pmatrix}$ and $\begin{pmatrix} 2\\\ 0 \\\ 0\end{pmatrix}$) then normalising it to get the unit normal vector, then scaling it to the area you've already computed. You should get $\vec A = \begin{pmatrix} 0 \\\ 4 \\\ 2\end{pmatrix}$. The required dot product, and hence the integral, is then $2$.