Solving a Cylindrical Curl Integral Question

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SUMMARY

The discussion focuses on solving a cylindrical curl integral involving the vector field \(\vec F\) and its curl, \(\nabla \times \vec F\). The user correctly identifies the curl using a determinant matrix and transforms the coordinates from Cartesian to cylindrical using \(x = r \cos(\theta)\), \(y = r \sin(\theta)\), and \(z = r\). The integral for the flux is established as \(\int_{0}^{2\pi}\int_{0}^{h} (r^{1.5} \sin(\theta) \cos(\theta) + r^{0.5} \sin(\theta) + \frac{1}{4}) dr d\theta\). The user seeks clarification on the correctness of their calculations and the appropriate value for \(h\) in the integral.

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Homework Statement



Question Attached

Homework Equations


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&amp;y&amp;z\\ \noalign{\medskip}{\frac {d}{{\it <br /> dx}}}&amp;{\frac {d}{{\it dy}}}&amp;{\frac {d}{{\it dz}}}\\ \noalign{\medskip}<br /> -z&amp;-x&amp;{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) <br /> \\ \noalign{\medskip}1/2\,\sqrt {r}\cos \left( \theta \right) <br /> \\ \noalign{\medskip}0\end {array} \right]

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

Thus finally, yielding the integral for the flux as.

\int _{0}^{2\,\pi}\!\int _{0}^{h}\!{r}^{ 1.5}\sin \left( \theta<br /> \right) \cos \left( \theta \right) +{r}^{ 0.5}\sin \left( \theta<br /> \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:
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except that you forgot the attachment ;)
 
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.
 

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