# Flux of curl

1. Apr 21, 2010

### Gameowner

1. The problem statement, all variables and given/known data

Question Attached

2. Relevant equations

3. The attempt at a solution

So here I'm attempting b), I know $$\nabla \times \vec F\$$ is the curl, which in this case is defined by the matrix

\left[ \begin {array}{ccc} x&y&z\\ \noalign{\medskip}{\frac {d}{{\it dx}}}&{\frac {d}{{\it dy}}}&{\frac {d}{{\it dz}}}\\ \noalign{\medskip} -z&-x&{y}^{2}\end {array} \right]

which gives me the vector $$\nabla \times \vec F\ = 2*yi+j-k$$

Since my given vector is a function of $$theta$$ and z, I apply the change of coordinates x=rcos(theta), y=rsin(theta), z=r which effectively changes the vector from

r(theta,z)=(sqrt(z)*cos(theta) , sqrt(z)/2*sin(theta) , z)

into

r(theta,r)=(sqrt(r)*cos(theta) , sqrt(r)/2*sin(theta) , r)

The equation of the integral being $$\int\int_S \nabla \times \vec F\cdot \hat n\, dS$$

so the $$\hat n\$$ is given by $$\hat n = \vec R_\theta \times \vec R_r$$ for downward pointing normal.

Rtheta
\left[ \begin {array}{c} -\sqrt {r}\sin \left( \theta \right) \\ \noalign{\medskip}1/2\,\sqrt {r}\cos \left( \theta \right) \\ \noalign{\medskip}0\end {array} \right]

Rr
\left[ \begin {array}{c} 1/2\,{\frac {\cos \left( \theta \right) }{ \sqrt {r}}}\\ \noalign{\medskip}1/4\,{\frac {\sin \left( \theta \right) }{\sqrt {r}}}\\ \noalign{\medskip}1\end {array} \right]

Thus taking the cross product and then yielding the normal

n=
\left[ \begin {array}{c} 1/2\,\sqrt {r}\cos \left( \theta \right) \\ \noalign{\medskip}\sqrt {r}\sin \left( \theta \right) \\ \noalign{\medskip}-1/4\end {array} \right]

Then I substitute the cylindrical coordinates x=rcos(theta), y=rsin(theta), z=r into my vector F to get.

\left[ \begin {array}{c} 2\, \left( r \right) \sin \left( \theta \right) \\ \noalign{\medskip}1\\ \noalign{\medskip}-1\end {array} \right]

Thus finally, yielding the integral for the flux as.

$$\int _{0}^{2\,\pi}\!\int _{0}^{h}\!{r}^{ 1.5}\sin \left( \theta \right) \cos \left( \theta \right) +{r}^{ 0.5}\sin \left( \theta \right) +1/4{dr}\,{d\theta}$$

Just want to know if everything I've done so far is correct(most of it were wild/educated guesses), and the domain for z is confusing me with the h>0, when I'm evaluating the integral, what would I put as h?

Thank you.

Last edited: Apr 21, 2010
2. Apr 21, 2010

### CompuChip

except that you forgot the attachment ;)

3. Apr 21, 2010

### Gameowner

Lol I'm so sorry, I was meant to click the add attachment button and went too fast and clicked the submit thread button too quick ><, editing it with all my working so far as we speak.