# Calculate the flux.

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• May 19th 2006, 01:49 AM
Bert
Calculate the flux.
Given is $\vec{v}=x\vec{u}_1+y\vec{u}_2+z\vec{u}_3$ and they ask me to calculate the flux if

$\left \{ \begin{array} \ z=x^2+y^2 \\ \ x^2+y^2 \leq 1 \end{array} \right$

Where $\vec{n}$ is in the up direction.

First I need to make a function who discripe the surface so $\left\{ \begin{array} \ x=x \\ y=y \\ z=x^2+y^2 \right \end{array}$ thus $\vec{F}=x\vec{u}_1 + y\vec{u}_2 + (x^2+y^2)\vec{u}_3$

and $\vec{F}_x=\vec{u}_1 +2x\vec{u}_3$ and $\vec{F}_y=\vec{u}_2 +2y\vec{u}_3$
If I calculate the cros product of the two I get $\vec{e}_3-2x\vec{e}_1-2y\vec{e}_2$ then $\left ( \begin{array} \ -2x \\ -2y \\ 1 \\ \end{array} \right ) \left ( \begin{array} \ x \\ y \\ z \\ \end{array} \right ) \right \begin{array} \ =-2x^2-2y^2+z \end{array} \left$

Then the integral $\int^{1}_{0} \int^{\sqrt{1-x^2}}_0-2x^2-2y^2+x^2+y^2 =\int^{1}_{0} \int^{\sqrt{1-x^2}}_0-x^2-y^2$
Is this methode oké? if I calculate the integral I get $\frac{\pi}{4}$ and not $\frac{\pi}{2}$

Who can help? Greets.
• May 19th 2006, 02:28 PM
TD!
You can also use Stokes' theorem, or calculate the flux like you did.

Perhaps this topic may be of assistance, I did it in both ways there. If you still can't get it, I'll look into your problem.
• May 22nd 2006, 04:53 AM
Bert
few question about this topic what do you mean with $curl\vec{v}$ do you mean the gradient?

then if you solve it wihout stokes then you don't use $(2y,xy-2xz,1-xz)$ any more why?

Greets.
• May 22nd 2006, 08:35 AM
TD!
The curl is a differential operator such as the gradient. I only used it there because that was part of the problem, see the initial post.

I'm not sure what you mean with your second question, the curl of v is actually used...
• May 22nd 2006, 09:31 AM
Bert
• May 22nd 2006, 10:00 AM
TD!
Read the next line: we're integrating over the circle in the xy-plane, so z = 0. Putting z = 0 in that first expression gives the second one of the scalar product - there I use it.
• May 22nd 2006, 10:48 AM
Bert
what is the value of $\nabla$

Greets.
• May 23rd 2006, 08:05 AM
TD!
It doesn't have a value, it's the nabla operator. It can be a gradient, a divergence or a curl - depening on how it is applied. Here, it is the cross product which gives the curl of the vector.

Edit: I just read your entire post: recheck your integral boundaries.
• May 25th 2006, 12:31 PM
Bert
can I rewrite $curl\vec{v}=\nabla (xyz+2,x,x^2z+y^2)$to $curl\vec{v}=(xyz+2,x,x^2z+y^2)*(xyz+2,x,x^2z+y^2)$

Greets.
• May 26th 2006, 02:42 AM
TD!
Euhm no. You have to take the curl, what you did was the dot product of the vector with itself, I think...
• May 26th 2006, 04:45 AM
Bert
of course I mean $
curl\vec{v}=(xyz+2,x,x^2z+y^2)\times(xyz+2,x,x^2z+ y^2)
$
thus the cross product not the dot. The problem was that I don't know who to write the cross in latex.

Can I rewrite it on this way? Greets.
• May 26th 2006, 04:49 AM
TD!
No, the curl (or the nabla operator on the vector via the cross product) doesn't mean the cross product of the vector with itself. Have you seen the curl yet? If not: do you even have to use it here? Check mathworld.com or wikipedia.com to see what the curl means :)