# Math Help - Divergence Theorem Help!

1. ## Divergence Theorem Help!

Here is the promblem

$F(x,y,z) = xz i + yz j + 3z^2 k$
$E\ is\ the\ solid\ bounded\ by\ the\ paraboloid\ z\ = x^2 + y^2\ and\ the\ plane\ z = 1$
$\int\int F \cdot dS = ?$

Divergence Theorem which i really don't understand well is $\int\int F \cdot dS = \int\int\int div\ F\ dV$

I went ahead and got the gradient vector which is $$ and then the function i got when i set up the triple integral is $8\int\int\int z dV$

and i guess all that is left for me to do is to set up and integrate. Math experts am im right up until this part, and if i am help me set up the triple integral. I'm assuming i have to switch to polar coordinates, i hope i don't have to switch to spherical or cylindrical coordinates, because i really don't understand that stuff very well!

2. Originally Posted by purplerain
Here is the promblem

$F(x,y,z) = xz i + yz j + 3z^2 k$
$E\ is\ the\ solid\ bounded\ by\ the\ paraboloid\ z\ = x^2 + y^2\ and\ the\ plane\ z = 1$
$\int\int F \cdot dS = ?$

Divergence Theorem which i really don't understand well is $\int\int F \cdot dS = \int\int\int div\ F\ dV$

I went ahead and got the gradient vector which is $$ and then the function i got when i set up the triple integral is $8\int\int\int z dV$

and i guess all that is left for me to do is to set up and integrate. Math experts am im right up until this part, and if i am help me set up the triple integral. I'm assuming i have to switch to polar coordinates, i hope i don't have to switch to spherical or cylindrical coordinates, because i really don't understand that stuff very well!
By the divergence theorem, the integral you seek is $\int \int \int \text{div}\bold F ~dV = \int_{-1}^1 \int_{- \sqrt {1 - x^2}}^{\sqrt{1 - x^2}} \int_{x^2 + y^2}^1 (z + z + 6z)~dz~dy~dx$ $= \int_0^{2 \pi} \int_0^1 \int_{r^2}^1 8z \cdot r ~dz~dr~d \theta$

3. Originally Posted by Jhevon
By the divergence theorem, the integral you seek is $\int \int \int \text{div}\bold F ~dV = \int_{-1}^1 \int_{- \sqrt {1 - x^2}}^{\sqrt{1 - x^2}} \int_{x^2 + y^2}^1 (z + z + 6z)~dz~dy~dx$ $= \int_0^{2 \pi} \int_0^1 \int_{r^2}^1 8r^2 \cdot r ~dz~dr~d \theta$
I tried evaluating it with the bounds you set up and i got 4.18879 which was marked wrong . Im kind of confused on how you got $8r^2$ when you set it up.

4. Originally Posted by purplerain
I tried evaluating it with the bounds you set up and i got 4.18879 which was marked wrong . Im kind of confused on how you got $8r^2$ when you set it up.
I'm sorry. I put $z = x^2 + y^2 = r^2$, but you should leave it as $z$. My bad. Try it again

and try to find the exact answer rather than a decimal approximation.

5. Originally Posted by Jhevon
I'm sorry. I put $z = x^2 + y^2 = r^2$, but you should leave it as $z$. My bad. Try it again

and try to find the exact answer rather than a decimal approximation.
Thanks, it worked!