Question

Veryify Gauss's theorem for the given three dimensional region D and vector field F.
F= (x,y,z)
D = {(x,y,z)|0<=z<=9-x^2-y^2}

Answer 1

Gauss's theorem says:
\[\int\limits_S \vec{V} \cdot d \vec{a} = \int\limits_V (\vec{\nabla} \cdot \vec{V}) dV\]
Therefore you need to do both of these integrals.

Answer 2

I am aware. I got the triple integral but I don't know how to set up the surface integral.

Answer 3

Oh okay. Well you need to parametrize your surface with two independent parameters:
\[\vec{\Phi}(x,y) = (x,y,9-x^2-y^2)\]
From there you take the derivative with respect to the two parametrizations and cross them:
\[\frac{\partial \vec{\Phi}}{\partial x} \times \frac{\partial \vec{\Phi}}{\partial y}\]
Then evaluate your vector field at phi:
\[\vec{F}(\vec{\Phi}(x,y))\]
Then you have:
\[\iint \vec{F}({\vec{\Phi}(x,y)}) \cdot \left( \frac{\partial \vec{\Phi}}{\partial x} \times \frac{\partial\vec{\Phi}}{\partial y}\right)dx dy\]
Now to find the range on x and y you need to set z to zero giving:
\[9 = x^2+y^2 \implies 0 \le x, y \le 3\]

Answer 4

And don't forget to check whether it is orientation preserving or reversing! Though I don't really recall how to check :/

Answer 5

Do you remember @zepdrix ? I know what an orientable surface is but that doesn't help with figuring out preserving/reversing...

Answer 6

haha no i dunno XD I think this math is a smidgin above my current level

Answer 7

From what I remember, the path needs to have the entire surface on the left side of the path which will show itself when I take the integral.

Answer 8

Also doesn't my surface need to be closed so I would have to add a disk of radius 3 at the bottom?

Answer 9

It is indeed, so you would need to add both parts. But for the other part its easy because you just have a parametrization of:
\[\vec{\Phi}(r,\phi) = (r \cos(\phi), r \sin(\phi),0)\]
As z = 0 for the disk.

Answer 10

\[0 \le r \le 3; 0 \le \phi \le 2 \pi\]

Answer 11

Yeah thats what I thought. Just wanted to be sure. Thank you

Answer 12

No problem!