I have surfaces with equations $$z=ax^2+by^2 \quad\text{and}\quad z=6-cx^2-dy^2.$$
I want to find the volume bounded between the these two surfaces (obviously paraboloids). I tried to use multiple integrals, but turns out to be impossible. I'm thinking how do I use some linear transformation of the coordinates to make it an easier integral? Any ideas?
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} V & \equiv \iiint_{\large\mathbb{R}^{3}}\bracks{z > ax^{2} + by^{2}} \bracks{z < 6 - cx^{2} - dy^{2}}\dd x\,\dd y\,\dd z \\[5mm] & = \iint_{\large\mathbb{R}^{2}}\int_{0}^{\infty} \bracks{ax^{2} + by^{2} < z < 6 - cx^{2} - dy^{2}}\dd z\,\dd x\,\dd y \\[5mm] & = \iint_{\large\mathbb{R}^{2}} \bracks{ax^{2} + by^{2} < 6 - cx^{2} - dy^{2}} \int_{ax^{2} + by^{2}}^{6 - cx^{2} - dy^{2}}\dd z\,\dd x\,\dd y \\[5mm] & = 6\iint_{\large\mathbb{R}^{2}} \bracks{{a + c \over 6}\,x^{2} + {b + d \over 6}\,y^{2} < 1} \pars{1 - {a + c \over 6}\,x^{2} - {b + d \over 6}\,y^{2}}\,\dd x\,\dd y \end{align}
\begin{align} V & = {36 \over \root{\pars{a + c}\pars{b + d}}} \iint_{\large\mathbb{R}^{2}}\bracks{x^{2} + y^{2} < 1} \pars{1 - x^{2} - y^{2}}\dd x\,\dd y \\[5mm] & \stackrel{\substack{2D\ Cyl. \\ Coord.}}{=}\,\,\, {36 \over \root{\pars{a + c}\pars{b + d}}} \int_{0}^{2\pi}\int_{0}^{1} \pars{1 - r^{2}}r\,\dd r\,\dd\phi = \bbx{{18\pi \over \root{\pars{a + c}\pars{b + d}}}} \end{align}