Consider the set
$$T = \{(x,y,z) \in \mathbb{R}^3:z^4\le x^2+y^2\le z^2, 0\le z\le 1\} $$
Compute the flux of the vector field $$ F = (x, y^2, \arctan(z)) $$ exiting through $\partial T $.
The standard way I was teached to solve this kind of problem is by means of the divergence theorem to transform the flux integral into a volume integral that is easier to solve:
$$\int_TF\cdot \hat ndS = \int_V\nabla \cdot FdV$$
I tried to solve the volume integral by using cylindrical coordinates that made the integration boundaries of $\rho, \theta$ and $z$ easy to obtain (respectively $z^4\ to\ z^2, 0\ to\ 2\pi, 0\ to\ 1$). However, I wasn't able to solve the integral because it gets too complex.
Does any one have suggestions? Is the route correct?
Best Answer
Your solid $T$ can be imagined by rotating the region in the $xz-$plane defined by $$\{(x,0,z):x\leq z \leq \sqrt{x},0\leq x \leq 1\}$$ about the $z-\text{axis}$. In cylindrical coordinates this can be described as $$T=\{(r\cos(\theta),r\sin(\theta),z)\in \mathbb{R}^3:r\leq z \leq \sqrt{r},0\leq r \leq 1,0\leq \theta < 2\pi\}$$ So with divergence theorem, $$\iint_{\partial{T}}(F\cdot n)\mathrm{d}S=\int_0^{2\pi}\int_0^1 \int_{r}^{\sqrt{r}}\left(1+2r\sin(\theta)+\frac{1}{z^2+1}\right)r\mathrm{d}z \mathrm{d}r \mathrm{d}\theta$$ Using standard integration techniques this evaluates to $-\frac{\pi^{2}}{2}+\frac{9\pi}{5}$.