I have a surface defined as $\displaystyle z(x^2+y^2)=4$. I have calculated the surface integral but now I just want to calculate the contribution from the disc z=4. How do i do this?
The surface integral was 33pi. In part (b), I had to calculate volume integral where V is the region defined by $\displaystyle 0<x^2+y^2<4/z$ and $\displaystyle 1<z<4$. This was 48pi.
Now I have to verify divegence theorem. Well, I need to show 33pi + contribution from z=4 + contribution from z=1 = 48pi.
Sorry if I was not clear enough.
In cylindrical coordinates, $\displaystyle x^2+ y^2= \frac{4}{z}$ becomes $\displaystyle z= \frac{4}{r^2}$. Of course, in order to have a closed figure, we must have the "top" and "bottom", z= 4, and z= 1.
The divergence theorem says that $\displaystyle \int\oint\vec{F}d\vec{S}= \int\int\int \nabla\cdot\vec{F} dV$.
If you want to find the volume of the figure, $\displaystyle \int\int\int dV$, using that, you must have a vector function, $\displaystyle \vec{F}$, such that $\displaystyle \nabla\cdot\vec{F}= 1$. What vector function did you use?
Of course, the top of the figure, z= 4, is the disk $\displaystyle x^2+y^2\le 1$ while the bottom of the figure, z= 1, is the disk $\displaystyle x^2+ y^2= 4$ which means that, to find the volume directly, you must integrate the constant 1 from z= 0 to z= 4 over the disk $\displaystyle x^2+ y^2= 1$ which is simply 1 times 4 times the area of that disk, $\displaystyle 4\pi$. Then, integrate the constant 1 from z= 0 to $\displaystyle z= \frac{4}{x^2+ y^2}= \frac{4}{r^2}$ (in cylindrical coordinates) for r= 1 to 2 (and $\displaystyle \theta$ from 0 to $\displaystyle 2\pi$, of course).