Show by divergence theorem $\int_SF.n\, dS=4\pi$ where $F=(x-z)i + (x^3+yz)j - 3xy^2k$

Question

Show by divergence theorem $$\int_SF.n\, dS=4\pi$$ where $F=(x-z)i + (x^3+yz)j - 3xy^2k$ and S is the surface of the cone $z=2-\sqrt{x^2+y^2}$ above the xy-plane.

$div\,F=1+z$.

$$\int_V\,div\,F\,dV=\int_V(1+z)\,dV$$

Edit:

Using cylindrical co-ordinates, $x= r \cos\theta; y=r \sin\theta; z=z$, $z \in [0,2-r], \theta\in [0,2\pi], r\in [0,2]$, we have

$$\int_{r=0}^{2}\!\int_{\theta=0}^{2\,\pi}\!\int_{z=0}^{2-r}\!(1+z)\,{\rm d}z \,{\rm d}\theta\,{\rm d}r=\frac{20\pi}{3} $$

Ans would be $4\pi$.....! Where is the mistake? Please suggest parametrization for the inverted cone and correct the limits of integration.

Answer 1

The limits of integration are correct, but you missed a factor $r$ due to the cylindrical coordinates, $$\int_{r=0}^{2}\!\int_{\theta=0}^{2\,\pi}\!\int_{z=0}^{2-r}\!(1+z)r\,{\rm d}z \,{\rm d}\theta\,{\rm d}r=|V|(1+\bar{z})=4\pi$$ where $|V|= \frac{8\pi}{3}$ is the volume of the cone and $\bar{z}=\frac{1}{2}$ is the $z$-coordinate of its centroid.

Answer 2

Hint

I'm surely rusty but, why not use cylindrical, so

$$\iiint_V(1+z)r\rm dr\rm d\theta\rm dz=\int_0^2\int_0^{2-z}\int_0^{2\pi}(1+z)r\rm d\theta \rm dr\rm dz $$, with $0\le z\le2, 0\le \theta\le2\pi$.

So...