1. ## Calculate the flux.

Given is $\displaystyle \vec{v}=x\vec{u}_1+y\vec{u}_2+z\vec{u}_3$ and they ask me to calculate the flux if

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

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

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

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

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

Who can help? Greets.

2. 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.

3. few question about this topic what do you mean with $\displaystyle curl\vec{v}$ do you mean the gradient?

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

Greets.

4. 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...

5. 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.

6. what is the value of $\displaystyle \nabla$

Greets.

7. 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.

8. can I rewrite $\displaystyle curl\vec{v}=\nabla (xyz+2,x,x^2z+y^2)$to $\displaystyle curl\vec{v}=(xyz+2,x,x^2z+y^2)*(xyz+2,x,x^2z+y^2)$

Greets.

9. Euhm no. You have to take the curl, what you did was the dot product of the vector with itself, I think...

10. of course I mean $\displaystyle curl\vec{v}=(xyz+2,x,x^2z+y^2)\times(xyz+2,x,x^2z+ y^2)$ 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.

11. 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