Compute a $2$D integral

Question

$$I=\iint_{D}\cfrac{dxdy}{\sqrt{1-\cfrac{{x}^{2}}{{a}^{2}}-\cfrac{{y}^{2}}{{b}^{2}}}\times\left( {x}^{2}+{y}^{2}+1-\cfrac{{x}^{2}}{{a}^{2}}-\cfrac{{y}^{2}}{{b}^{2}}\right)^{\cfrac{3}{2}}}$$ where $$D=\left\{(x,y);\cfrac{{x}^{2}}{{a}^{2}}+\cfrac{{y}^{2}}{{b}^{2}}\leq 1,x\geq 0,y\geq 0\right\}.$$

How to compute then? What I can image is just the radial transformation: $x=ar\cos t, y=br\sin t$.

I get $$\int_0^1 abr dr\int_0^{\pi/2} \cfrac{dt}{\sqrt{1-r^2}(a^2r^2\cos^2t+b^2r^2\sin^2t+1-r^2)^{3/2}}.$$ — XLDD, Oct 14 '13 at 00:35
@MichaelAlbanese There is a limit of characters, less than 150 words... — XLDD, Oct 14 '13 at 00:38
I think you are mistaken. You can have more than $150$ words in a question. — Michael Albanese, Oct 14 '13 at 00:40
You should integrate $r$ first by a substitution of variable $u = \frac{r^2}{1-r^2}$. — achille hui, Oct 14 '13 at 01:54

score 1 · Accepted Answer · answered Oct 14 '13 at 01:59

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\expo}{{\rm e}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\pp}{{\cal P}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}$ $\large\it Hint:$

\begin{align} I &= {1 \over 2}\,\verts{ab} \int_{0}^{\pi/2}{\rm d}\theta\int_{0}^{1}{\rm d}r\, {1 \over \sqrt{1 - r\,}\, \braces{\vphantom{\LARGE A}\bracks{\vphantom{\Large A}% a^{2}\cos^{2}\pars{\theta} + b^{2}\sin^{2}\pars{\theta} - 1}r + 1}^{3/2}} \end{align}

I can't see how the rightmost integral is less monstruous than the original one... — DonAntonio, Oct 14 '13 at 03:50

score 0 · Answer 2 · answered Oct 14 '13 at 10:06

Let $$\Delta(\theta) = a^2 \cos^2(\theta) + b^2\sin^2(\theta)$$ In terms of the parametrization $(x,y) = (ar\cos\theta,br\sin\theta)$, the integral $\mathscr{I}$ we want can be rewritten as

$$\begin{align} \mathscr{I} = & ab \int_0^{\frac{\pi}{2}} d\theta \int_0^1 \frac{rdr}{\sqrt{1-r^2}\sqrt{\Delta(\theta) r^2 + 1 - r^2}^3}\\ = & \frac{ab}{2}\int_0^{\frac{\pi}{2}} d\theta \int_0^1 \frac{dr^2}{(1-r^2)^2}\frac{1}{\sqrt{\Delta(\theta) \frac{r^2}{1-r^2} + 1}^3} \end{align}$$

Let $u = \frac{r^2}{1-r^2}$, $v = \Delta(\theta) u + 1$ and $t = \tan\theta$. Notice $\displaystyle du = \frac{dr^2}{(1-r^2)^2} $, we have

$$\mathscr{I} = \frac{ab}{2}\int_0^{\frac{\pi}{2}} d\theta \int_0^{\infty} \frac{du}{\sqrt{\Delta(\theta) u + 1}^3} = \frac{ab}{2}\int_0^{\frac{\pi}{2}} \frac{d\theta}{\Delta(\theta)} \int_1^{\infty} \frac{dv}{\sqrt{v}^3}\\ = ab \int_0^{\frac{\pi}{2}} \frac{d\theta}{a^2\cos^2(\theta) + b^2\sin^2(\theta)} = ab \int_0^{\infty} \frac{dt}{a^2 + b^2 t^2} = \frac{\pi}{2} $$

@XLDD, it is $\frac{\pi}{2}$. When $a = b = R$, you can use wolframalpha to evaluate the $r$ part of the integral $I$ specified in Felix's answer. You will get $$\int_0^1\frac{dr}{\sqrt{(1-r)((R^2-1)r+1)^3}} = \frac{2}{R^2}$$ Combine it with the factor $\frac{R^2}{2}\frac{\pi}{2}$, you get $I = \frac{\pi}{2}$, a number independent of $R$. — achille hui, Oct 30 '13 at 03:20

Compute a $2$D integral

2 Answers2