Question

Sketch the solid described by x^2 + y2^ ≤ z ≤ 1. Use the Divergence Theorem to evaluate the surface integral over the boundary of that solid of the vector field F(x,y,z)=yi+zJ+xzk

Answer 1

the solid is
x^2 + y^2 ≤ z ≤ 1
and not
x^2+y^2=z; 0 ≤ z ≤ 1
???
never seen that notation befor

Answer 2

Same, what is the solid exactly? I think that is typed wrong.

Answer 3

it is a paraboloid cut off by z=1, I would say.

Answer 4

whatever, I'm just gonna do what I think it should be:\[r^2\le z\le1\]\[0\le r\le1\]\[0\le\theta\le2\pi\]\[div\overrightarrow{F}=1\]\[\int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{r^2}rdzdrd\theta=\frac{\pi}{2}\]sorry if it's wrong...

Answer 5

careful; div F = x

Answer 6

d'oh!
retrying...

Answer 7

\[r^2\le z\le1\]\[0\le r\le1\]\[0\le\theta\le2\pi\]\[div\overrightarrow{F}=x\]\[\int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{r^2}r^2\cos\theta dzdrd\theta=0\]I guess...
did I mess up again?

Answer 8

i got 0 too

Answer 9

well that's what we'll go with then :)

Answer 10

It is zero, because the solid is symmetric in x.

Answer 11

Yeah, I can see that, but I'm always suspicious of things like that.