$\int_{\mathbb{R}^2} \delta(E-ax-by) x^2 dx $ I am wondering how we have to integrate $\int_{\mathbb{R}^2} \delta(E-ax^2-by^2) x^2 dxdy.$ I am not familiar with this kind of delta distribution (depending on two coordinates), so I was wondering if there is a standard trick to evaluate this integral?
 A: Use Ruffini Theorem in order to integrate first with respect to $dx$ and then with repect to $dy$, that is,
$\begin{equation*}
\int_{\mathbb{R}^2} \delta(E-ax-by)x^2 dx dy = \int_{\mathbb{R}} \Big( \int_{\mathbb{R}} \delta(E-ax^2-by^2)x^2 dx \Big) dy
\end{equation*}$
Now you can think the delta function as a function of $x$ only, taking $y$ constant (true by computing the first integral); use the following property of the delta function:
$\begin{equation*}
\delta( f(x) ) = \sum_{x_0 \text{ zero of $f$}} \frac{\delta(x-x_0)}{|f'(x_0)|}
\end{equation*}$
P.D. By using this method you must deal with square roots: in order to avoid them you can use a change of variables: try changing
$\begin{equation*}
 \phi : \mathbb{R}^2 \longrightarrow (0,+\infty) \times (0, 2 \pi)
\end{equation*}$
given by $\phi(x,y) = (\frac{1}{\sqrt{a}}·r· \cos(t), \frac{1}{\sqrt{b}}·r· \sin(t))$, $r$ varying in $(0,+\infty)$, and $t$ in $(0, 2 \pi)$. (Don't forget the Jacobian!)
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\color{#66f}{\large\int_{{\mathbb R}^{2}}%
\delta\pars{E - ax^{2} - by^{2}}x^{2}\,\dd x\,\dd y}
\\[5mm]&=\int_{-\infty}^{\infty}\dd x\,x^{2}\,\Theta\pars{E - ax^{2} \over b}
\times
\\&\int_{-\infty}^{\infty}
{\delta\pars{y + \root{\bracks{E - ax^{2}}/b}}
+ \delta\pars{y - \root{\bracks{E - ax^{2}}/b}} \over \verts{2by}}
\,\dd y
\\[5mm]&={1 \over \verts{b}}
\int_{-\infty}^{\infty}\Theta\pars{E - ax^{2} \over b}x^{2}\,
{\dd x \over \root{\pars{E - ax^{2}}/b}}
\end{align}

There are several possibilities according to the signs of $\ds{E, a}$ and $\ds{b}$.

I'll perform an evaluation in the case $\ds{E >0\,,\ a>0\,,b>0}$:
\begin{align}&\color{#66f}{\large\int_{{\mathbb R}^{2}}%
\delta\pars{E - ax^{2} - by^{2}}x^{2}\,\dd x\,\dd y}
={1 \over \root{b}}\ \overbrace{%
\int_{-\root{E/a}}^{\root{E/a}}{x^{2}\,\dd x \over \root{E - ax^{2}}}}
^{\ds{\color{#c00000}{x \equiv \root{E \over a}\sin\pars{\theta}}}}
\\[5mm]&={2 \over \root{b}}
\int_{0}^{\pi/2}{\pars{E/a}\sin^{2}\pars{\theta} \over \root{E}\cos\pars{\theta}}\,
\root{E \over a}\,\cos\pars{\theta}\,\dd\theta
={2E \over a^{3/2}b^{1/2}}\
\overbrace{\int_{0}^{\pi/2}\sin^{2}\pars{\theta}\,\dd\theta}^{\ds{=\ \color{#c00000}{\pi \over 4}}}
\end{align}

\begin{align}&\color{#66f}{\large\int_{{\mathbb R}^{2}}%
\delta\pars{E - ax^{2} - by^{2}}x^{2}\,\dd x\,\dd y}
=\color{#66f}{\large{\pi \over 2}\,{E \over a^{3/2}b^{1/2}}}\,,\qquad
E>0\,,\ a>0\,,\ b>0
\end{align}

