# Thread: flux using divergence theorem

1. ## flux using divergence theorem

Let W be the solid bounded by the paraboloid $\displaystyle x = y^2 + z^2$ and the plane x = 9. Let $\displaystyle \vec{F} = 5x\vec{i} + y\vec{j} + z\vec{k}$. I know the divF is 7 here.. but after that I will need to multiply this by the volume of the region bounded above to get the flux.. am I right?

2. Originally Posted by TheRekz
Let W be the solid bounded by the paraboloid $\displaystyle x = y^2 + z^2$ and the plane x = 9. Let $\displaystyle \vec{F} = 5x\vec{i} + y\vec{j} + z\vec{k}$. I know that $\displaystyle div(\vec{F})=7$ here.. but after that I will need to multiply this by the volume of the region bounded above to get the flux.. am I right?
Yes, you want to integrate $\displaystyle \int\int\int 7\,dV$.

I would use cylindrical coordinates (note that I'm using x where one would traditionally use z):
$\displaystyle x=x$
$\displaystyle r^2=y^2+z^2$
$\displaystyle dV=r\,dx\,dr\,d\theta$

So your volume is bounded by $\displaystyle x=r^2$ and $\displaystyle x=9$

$\displaystyle \theta=0..2\pi$
$\displaystyle r=0..3$
$\displaystyle x=r^2..9$

$\displaystyle 7\int_0^{2\pi}\int_0^3\int_{r^2}^9 r\,dx\,dr\,d\theta$

Spoiler:
Inner integral: $\displaystyle \left[rx\right]_{r^2}^9 = 9r-r^3$

Now you have $\displaystyle 7\int_0^{2\pi}\int_0^3 9r-r^3\,dr\,d\theta$

Middle integral: $\displaystyle \left[\frac{9r^2}{2}-\frac{r^4}{4}\right]_0^3 = \frac{81}{2}-\frac{81}{4} = \frac{81}{4}$

Now you have $\displaystyle \frac{7\cdot 81}{4}\int_0^{2\pi}\,d\theta = \frac{2\pi\cdot 7\cdot 81}{4} = \frac{567\pi}{2}$

PS: The closing tag for LaTeX is "[/tex]", not "[\math]".

3. hey thanks for that, what if it asks:

Let S1 be the paraboloid surface oriented in the negative x direction, what is the flux through S1?