Exercise
Consider the region $$ D=\{(x, y, z)\in\mathbb{R}^3\,|\, x^2+y^2\leq9, y^2+z^2\leq25\}. $$
1. Write down a parametrization of the boundary $∂D$ of $D$.
2. Compute the outward pointing unit normal field n of the boundary $∂D$.
3. Compute the outward flux of the vector field $$ F(x,y,z)=\Big(x\sqrt{25-y^2},\,\,x\sqrt{25-z^2},\,\,y\sqrt{9-x^2}\Big) $$ through the closed surface $∂D$.
Solution.
1. The region $D$ consists of that portion of a "vertical" cylinder (whose axis coincides with the $z$ axis) which is contained into another "horizontal" cylinder (whose axis coincides with the $x$ axis). We can split the boundary $∂D$ of $D$ into three parts, and provide three different parametrizations: $$ \mathbb{R}^2\supseteq\{u^2+v^2\leq9\}\ni(u, v)\longmapsto\Big(u, v, \sqrt{25-v^2}\Big) $$ for the upper "cover"; $$ \mathbb{R}^2\supseteq\{u^2+v^2\leq9\}\ni(u, v)\longmapsto\Big(u, v, -\sqrt{25-v^2}\Big) $$ for the lower "cover"; $$ \mathbb{R}^2\supseteq\{0\leq\theta\leq2\pi,\,\,9\sin^2(\theta)+z^2\leq25\}\ni(\theta, z)\longmapsto\Big(3\cos(\theta),\,\,3\sin(\theta),\,\, z\Big) $$ for the side (using cylindrical coordinates).
2. It is easy to see that we have: $$ n_1=\Big(0, \frac{v}{\sqrt{25-v^2}}, 1\Big) $$
$$ n_1=\Big(0, -\frac{v}{\sqrt{25-v^2}}, 1\Big) $$
$$ n_3=(\cos(\theta), \sin(\theta), 0). $$ I am struggling with computing the outward flux of the vector field $F$. I tried both using the definition of outward flux of a vector field through a surface and using the divergence theorem and cylindrical coordinates (which quite looks as the right way to approach this third question, as the divergence of the vector field $F$ is quite simple). Nevertheless, I cannot get anywhere. Can anybody provide some help?