finding outward flux of the field

**Lafexlos** · July 17th 2011, 12:27 PM

hi.

Work out the outward flux of the field $\vec F=y\hat i+x\hat j + z\hat k$ across the boundary of the volumetric region bounded bty $z^2=x^2+y^2$ and $z=1$ . Verify your result using divergence theorem.

that's the question. i tried solve like this;
$\iiint_V \nabla\cdot F dV$

$\nabla\cdot F=1$ hence volume part of divergence theorem becomes;

$\iiint_V dV$ where our volumetric region is $z^2=x^2+y^2$

after this i couldn't understand, which coordinate system should i work and what should i write as bounds of integrals.

as i look to surface part of the divergence theorem, i used these substitutions;
$\iint_D \left \vec F \cdot \hat n \right d\sigma$ ,,,, $\hat n= \frac{\nabla f}{|\nabla f|}$ ,,,, $d\sigma = \frac{|\nabla f|}{|\nabla f \cdot \hat p|}dA$
where;
$\vec F = y\hat i + x\hat j + z\hat k$ and $f=x^2+y^2-z^2$

these substitutions didn't give me an answer so i need help on both sides of divergence theory.

edit: btw, sorry for grammer mistakes if there is any. my english grammer knowledge isn't good at all.

**ojones** · July 17th 2011, 06:27 PM

Give the details of your calculations; maybe I can spot the mistake.

**Lafexlos** · July 17th 2011, 08:35 PM

for the volumetric part, those all i could do.

for the other part;
$\nabla f = 2x\hat i + 2y\hat j - 2z\hat k$
$|\nabla f \cdot \hat p| = 2z$ and $|\nabla f|$ 's are cancelling each other because they are scalar.

$\iint_D \frac {2xy+2xy-2z^2}{2z} dA = \iint_D \frac {2xy-z^2}{z}dA$

from volumetric region, i wrote $z=\sqrt {x^2+ y^2}$

$\iint_D \frac {2xy-x^2-y^2}{\sqrt {x^2+y^2}}dA$

from this i couldn't wrote bounds of intergrals and didn't understand which coordinate system should i work at..

**ojones** · July 18th 2011, 01:07 PM

You haven't given enough details. There are two surfaces over which you must calculate the flux; part of the plane $z=1$ and the side of the cone $z^2=x^2+y^2$ . Give your answers to the following:

1. What are the normals to these surfaces?

2. What is the surface integral in the case of the plane?

3. What does the surface integral for the cone look like in terms of Cartesian coordinates $x$ and $y$ ?

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