I have to verify Gauss' Divergence Theorem for $$\bar F(x,y,z)=xy^2 \hat i+yz^2\hat j+zx^2\hat k$$ for the region $$R:y^2+z^2\le x^2;\;\;0\le x\le4$$
Now, $\operatorname{div}\bar F=x^2+y^2+z^2$. I don't understand what type of parametrization will make calculating the triple integral easy. I tried $\bar r=x\hat i+x\cos\theta\hat j+x\sin\theta\hat k$ where $0\le \theta \le 2\pi$. But then, $\operatorname{div}\bar F =r^2$. Would then $$\iiint_R \operatorname{div}\bar F\;dV=\int_0^4\int_0^{2\pi}\int_0^{\sqrt2x}r^2\;rdr\;d\theta\;dx$$ be correct?
I got it!
The solid we're looking for is an inverted cone. As $x$ increases, we have disks with $y^2+z^2\le x^2$, implying the radius of the disk formed at height $x$ is $x$ itself. Hence, the slant surface of the cone can be parametrized as $\bar r=x\hat i+xcos\theta\hat j+xsin\theta\hat k$. Let the slant surface be $S_1$. The normal vector $\bar n=-r_x \;\text{x}\;r_\theta =-x\hat i+y\hat j+ z\hat k$ Then $\iint_{S_1}\bar F\cdot \bar n \;dS$ can be calculated.
Let surface $S_2$ be the base of the cone. It can be parametrized as $y^2+z^2\le r^2\;;\;0\le r\le4$ giving $y=rcos\theta,\;z=rsin\theta$ and $\bar n=\hat i$. Then $\iint_{S_2}\bar F\cdot \bar n\;dS $ can be calculated.
And $\iint_{R}\bar F\cdot \bar n \;dS $=$\iint_{S_1}\bar F\cdot \bar n\;dS $ +$\iint_{S_2}\bar F\cdot \bar n \;dS$
Finally, $\text{div}\bar F=x^2+r^2$. So, $\iiint_R\text{div}\bar F \;dV=\int_0^4\int_0^{2\pi}\int_0^x(x^2+r^2)\;rdr\;d\theta\;dx$
Thus it can be verified.