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Consider an integral of the type

$$\int_{\mathbb{R}^2} \mathrm{d}x\mathrm{dy}\delta(x^2+y^2-a)$$

Where $a\in\mathbb{R}_+$. My first approach would be to make a substitution $x^2= s$, so the integral becomes

$$ \int_{\mathbb{R}^2} \mathrm{d}s\mathrm{d}y\delta(s+y^2-a)\frac{1}{2\sqrt{s}}$$

So I would instinctively write

$$\int_{\mathbb{R}}\mathrm{d}y\frac{1}{2\sqrt{a-y^2}}$$

but this is obviously wrong as the integrand is real only for $y\in[-\sqrt{a},\sqrt{a}]$

So then I would write

$$\int_{-\sqrt{a}}^\sqrt{a}\mathrm{d}y\frac{1}{2\sqrt{a-y^2}}$$

But this seems somewhat ad hoc and unjustified. How can I make sense of this? Please forgive the lack of rigor in treating $\delta$, this question arises from a physics problem, but it seemed more suitable here than in physics SE.

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First of all, I would use polar coordinates for this type of problem: $$ \int_{\mathbb{R}^2} \mathrm{d}x\mathrm{dy}\delta(x^2+y^2-a)=2\pi\int_0^\infty dr\ r\delta(r^2-a)=2\pi\int_0^\infty dr\ r\frac{\delta(r-\sqrt{a})}{2\sqrt{a}}=\pi\ , $$ result valid for any $a>0$. You cannot make the substitution $x^2=s$, as the $x$-limits of integration are $-\infty$ and $+\infty$ which are both mapped to $+\infty$ upon this substitution. You wouldn't use the substitution $x^2=t$ to evaluate the single integral $$ \int_\mathbb{R}dx\frac{1}{x^2+a^2}\ , $$ would you? At least, not before splitting the integral as $\int_\mathbb{R}=\int_0^\infty+\int_{-\infty}^0$...

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  • $\begingroup$ That was stupid of me. I should have thought of polar coordinates, thank you for the answer. $\endgroup$ – user438666 Nov 11 '17 at 18:31

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