Question

Set up a double integral for calculating the flux of the vector field F⃗ (x,y,z)=xi⃗ +yj⃗ through the open-ended circular cylinder of radius 8 and height 9 with its base on the xy-plane and centered about the positive z-axis, oriented away from the z-axis. If necessary, enter θ as theta.

Answer 1

Parameterize the cylinder (call it $S$) by

$\vec s(\theta,z)=\sqrt8\cos\theta\,\vec\imath+\sqrt8\sin\theta\,\vec\jmath+z\,\vec k$

with $0\le\theta\le2\pi$ and $0\le z\le9$. Then

$\vec F(x(\theta,z),y(\theta,z),z(\theta,z))=\sqrt8\cos\theta\,\vec\imath+\sqrt8\sin\theta\,\vec\jmath$

Take the normal vector to $S$ to be

$\vec s_\theta\times\vec s_z=\sqrt8\cos\theta\,\vec\imath+\sqrt8\sin\theta\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(\theta,z),y(\theta,z),z(\theta,z))\cdot(\vec s_\theta\times\vec s_z)\,\mathrm dz\,\mathrm d\theta$

$=\displaystyle\int_0^{2\pi}\int_0^98(\cos^2\theta+\sin^2\theta)\,\mathrm dz\,\mathrm d\theta=\boxed{144\pi}$