# Computing the flux of the vector field F

• Apr 13th 2013, 10:32 AM
SA2392
Computing the flux of the vector field F
Compute the flux of the vector field F = 5yi + 2j - 5xzk through the surface S, which is the surface y = x² + z², with x² + z² ≤ 9, oriented in the positive y-direction.

I understand that F = <5y, 2, -5xz>, but how do I find dA? and would the bounds for x and z be [0, 3]?

• Apr 13th 2013, 11:39 AM
HallsofIvy
Re: Computing the flux of the vector field F
I would change to polar coordinates in the xz-plane. That is, $x= r cos(\theta)$, $z= r sin(\theta)$, $y= x^2+ z^2= r^2$ with $r\le 3$
Now the "position vector" for any point on that paraboid would be $\vec{p}= x\vec{i}+ y\vec{j}+ z\vec{k}= r cos(\theta)\vec{i}+ r^2\vec{y}+ r sin(\theta)\vec{k}$. The derivatives are $\vec{p}_r= cos(\theta)\vec{i}+ 2r\vec{j}+ sin(\theta)\vec{k}$ and $\vec{p}_\theta= -r sin(\theta)\vec{i}+ r cos(\theta)\vec{k}$.

The cross product of those vectors is
$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ cos(\theta) & 2r & sin(\theta) \\ -r sin(\theta) & 0 & r cos(\theta)\end{array}\right|= 2r^2cos(\theta)\vec{i}- r\vec{j}+ 2r^2sin(\theta)\vec{k}$
so the "vector differential of surface area" is
$d\vec{S}= (2r^2cos(\theta)\vec{i}- r\vec{j}+ 2r^2sin(\theta))drd\theta$

But since we want "oriented in the positive y direction" we need to multiply throught by -1 to make the y-component positive:
$d\vec{S}= (-2r^2cos(\theta)\vec{i}+ r\vec{j}- 2r^2sin(\theta))drd\theta$
(That is the same as swapping the two derivative vectors in the cross product.)

Since $r\le 3$, and there is circular symmetry on the xz-plane, the limits of integration are r= 0 to 3 and $\theta= 0$ to $2\pi$.