Calculate the integral $ \int \limits_{V}\left(x^{2}+y^{2}\right) d(x, y, z) $, where $ V \subset \mathbb{ R}^{3} $ is limited by the areas $ x^{2}+y^{2}= $ $ 2 z $ and $ z=2 $.
One can see that the problem is symmetrical and $f(x,y)\geq 0$. So I can just focus on the space with $x\leq 0$ and $z\geq 0$ .) and later multiply the volume by 2. I want to solve this problem with cylindrical and cartesain coordinates:
Cylindrical coordiantes:
$ \left(\begin{matrix}x \\ y \\z \end{matrix}\right)=\left(\begin{matrix}r \cdot \cos (\phi) \\ r \cdot \sin (\phi)\\ z\end{matrix}\right) $
Boundaries:
$(1)\, 0\leq z \leq 2$.
$(2)\, -\frac{\pi}{2}\leq \phi \leq \frac{\pi}{2}$
$(3)\,x^2+y^2=r^2=2z\Rightarrow r=\sqrt{2z}$ because I can assume (symmetry) $r\neq -\sqrt{2z}$.
I would now do the following intgration:
$\int \limits_{V}\left(x^{2}+y^{2}\right) d(x, y, z)= 2 \int \limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int \limits_{0}^{2}\left( \int \limits_{0}^{\sqrt{2z}} r^{2} r\, dr\right)\, dz\, d\phi =2 \int \limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int \limits_{0}^{2}\left( \int \limits_{0}^{\sqrt{2z}} r^{3} \, dr \right)\, dz\, d\phi=2 \int \limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int \limits_{0}^{2}z^2\, dz\, d\phi=2 \int \limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{8}{3}\, d\phi=\dfrac{16\pi}{3} $
Would you agree with my result?
Cartesian coordinates:
Boundaries:
$(1)\, 0\leq z \leq 2$.
$(2)\, x^2+y^2=2z\Leftrightarrow x=\sqrt{2z-y^2}$ I assume $x\neq -\sqrt{2z-y^2}$ because of the symmetry. So $0\leq x\leq \sqrt{2z-y^2}$
$(3)$ How do I determine the boundaries of $y$?