Using divergence theorem, find flux of vector field ${\bf F}=\left(x\left(z - 1\right),y\left(z - 1\right),xy\right)$ on the area determined with inequalities $$ \left(z-1\right)^2\le 4-x^2-y^2,\quad \left(z-1\right)^2\ge x^2+y^2,\quad z\ge1 $$
- This comes down to calculating the integral $2\displaystyle\iiint_{V}\left(z-1\right)\,\mathrm{d}V$. Then, what is left is to find bounds.
- When I set $z=0$, in the $OXY$ plane I get $x^2+y^2\le 1$ and $x^2+y^2\le3$, so it is enough to watch circle with radius $\sqrt3$ and center at origin.
- After cylindrical coordinates I have $$ \left\{\begin{array}{rcccl} {\displaystyle 0} & {\displaystyle \le} & {\displaystyle \phi} & {\displaystyle \le} & {\displaystyle 2\pi} \\[0.5mm] {\displaystyle 0} & {\displaystyle \le} & {\displaystyle r} & {\displaystyle \le} & {\displaystyle \sqrt{3}} \\[0.5mm] {\displaystyle r + 1} & {\displaystyle \le} & {\displaystyle z} & {\displaystyle \le} & {\displaystyle 1 + \sqrt{4 - r^{2}}} \end{array}\right. $$ And then I would integrate in order ${\rm d}z, {\rm d}r, {\rm d}\phi$.
Is this correct?
Everything seems correct except bound of $r$.
This is a closed surface enclosed by an inverted cone along $z-$axis with vertex at $(0,0,1)$ and a sphere of radius $2$ with center at $(0,0,1)$. Radius of resulting surface is maximum where they intersect which we can find by equating them,
Sphere: $(z-1)^2\le 4-r^2$, Cone: $(z-1)^2\ge r^2$.
At intersection, $4-r^2 = r^2 \implies r = \sqrt2$
So bounds of $r$ is $0 \leq r \leq \sqrt2$.