Question

Calculate the flux of the vector field f⃗ (x,y,z)=cos(x2+y2)k⃗ through the disk x2+y2≤9 oriented upward in the plane z=4.

Answer 1

Parameterize the disk $S$ by

$\mathbf r(s,t)=\begin{cases}x(s,t)=s\cos t\\y(s,t)=s\sin t\\z(s,t)=4\end{cases}$

where $0\le s\le3$ and $0\le t\le2\pi$. Call this parameterized region $T$. Then

$\mathbf r_s\times\mathbf r_t=(\cos t\,\mathbf i+\sin t\,\mathbf j)\times(-s\sin t\,\mathbf i+s\cos t\,\mathbf j)=s\,\mathbf k$

The flux over the disk is given by the surface integral

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iint_T\mathbf f(x(s,t),y(s,t),z(s,t))\cdot(\mathbf r_s(s,t)\times\mathbf r_t(s,t))\,\mathrm ds\,\mathrm dt$
$=\displaystyle\int_{t=0}^{t=2\pi}\int_{s=0}^{s=3}(\cos(s^2)\,\mathbf k)\cdot(s\,\mathbf k)\,\mathrm ds\,\mathrm dt$
$=\displaystyle2\pi\int_{s=0}^{s=3}s\cos(s^2)\,\mathrm ds$
$=\displaystyle\pi\int_{\sigma=0}^{\sigma=9}\cos\sigma\,\mathrm d\sigma$

(where we take $\sigma=s^2$)

$=\pi\sin 9$