$$ \iiint z \, dV $$
over region D. D is bounded by by the paraboloid $z=4-x^2-4y^2 $ and the $xy$-plane. Use $x=2u\cos v ,y=u\sin v,z=w$
I found the Jacobian $J(u,v,w)=2u $.
I substituted $x=2u\cos v ,y=u\sin v,z=w$ in $z=4-x^2-4y^2 $ and got $$z=4-4u^2 $$
I understand that z axis' lower limit is 0 (because of the xy plane) and $z=4-4u^2 $. Other than that, I have no idea how I can get the limits of $du$ and $dv$. I'd appreciate if someone shows me step by step how we can get those limits because I'm really confused at this point.
thanks
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \iiint_{\mrm{D}}z\,\dd^{3}\vec{r} & = \int_{0}^{4}z\iint_{x^{2}\ +\ 4y^{2}\ <\ 4\ -\ z} \,\dd x\,\dd y\,\dd z = {1 \over 2}\int_{0}^{4}z\iint_{x^{2}\ +\ y^{2}\ <\ 4\ -\ z} \,\dd x\,\dd y\,\dd z \\[5mm] & = {1 \over 2}\int_{0}^{4}z\int_{0}^{2\pi}\int_{0}^{\root{4 - z}}\rho \,\dd\rho\,\dd\phi\,\dd z = {1 \over 2}\int_{0}^{4}z\bracks{2\pi\,{\pars{\root{4 - z}}^{2} \over 2}}\,\dd z \\[5mm] & = {1 \over 2}\,\pi\int_{0}^{4}\pars{4z - z^{2}}\,\dd z = \bbox[10px,#ffe,border:1px dotted navy]{\ds{{16 \over 3}\,\pi}} \approx 16.7552 \end{align}