1. ## divergence theorem

Can I use the divergence theorem to calculate the following:

$\vec{G} = -x^3\vec{i} - y^3\vec{j} - z^3\vec{k}$

Find the flux of the vector field out of the closed surface S. S is the boundary of the solid region W. W is the upper hemisphere of radius 3 centered at the origin.

well the divG I got is $-3x^2 -3y^2 -3z^2$, which is $-3(x^2 + y^2 + z^2)$ and this translates to $-3r^2$ am I right? do I just then multiply this by the volume, which is $18\pi$ in this case?

2. Originally Posted by TheRekz
Can I use the divergence theorem to calculate the following:

$\vec{G} = -x^3\vec{i} - y^3\vec{j} - z^3\vec{k}$

Find the flux of the vector field out of the closed surface S. S is the boundary of the solid region W. W is the upper hemisphere of radius 3 centered at the origin.

well the divG I got is $-3x^2 -3y^2 -3z^2$, which is $-3(x^2 + y^2 + z^2)$ and this translates to $-3r^2$ am I right? do I just then multiply this by the volume, which is $18\pi$ in this case?
You do know what the divergence theorem says, don't you?

$\int\int \vec{G}\cdot d\vec{S}= \int\int\int \nabla\cdot \vec{G} dV$
Since $\nabla\cdot \vec{G}$ is not constant in that solid, no, you do not "multiply by the volume", you integrate.

3. Originally Posted by HallsofIvy
You do know what the divergence theorem says, don't you?

$\int\int \vec{G}\cdot d\vec{S}= \int\int\int \nabla\cdot \vec{G} dV$
Since $\nabla\cdot \vec{G}$ is not constant in that solid, no, you do not "multiply by the volume", you integrate.
So I would have to translate that to a spherical coordinate first and integrate it?

4. Originally Posted by TheRekz
So I would have to translate that to a spherical coordinate first and integrate it?

5. Originally Posted by HallsofIvy
You do know what the divergence theorem says, don't you?

$\int\int \vec{G}\cdot d\vec{S}= \int\int\int \nabla\cdot \vec{G} dV$
Since $\nabla\cdot \vec{G}$ is not constant in that solid, no, you do not "multiply by the volume", you integrate.
$-3 \int_0^{2\pi}\int_0^{\pi/2}\int_0^3 \rho^2 sin \phi d\rho d\phi d\theta$

and the final result I have is -54pi... and it doesn't match with what the computer says

6. Originally Posted by TheRekz
$-3 \int_0^{2\pi}\int_0^{\pi/2}\int_0^3 \rho^2 sin \phi d\rho d\phi d\theta$

and the final result I have is -54pi... and it doesn't match with what the computer says
The answer is given by $-3 \int_0^{2\pi}\int_0^{\pi/2}\int_0^3 \rho^{\color{red}4} sin \phi d\rho d\phi d\theta$.

7. may I ask why it's to the power of 4?

8. Originally Posted by TheRekz
may I ask why it's to the power of 4?
Instead of posting so soon (within a minute of my post!) you should have spent time thinking about what a posted.

Do you know what dV is in spherical coordinates? Do you recall that the thing you're integrating (the divegrence) is $-3r^2$? Think carefully about these two questions.

9. thanks