Caculate $\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$, with D:$\frac{x^2}{2}+y^2\leq1$

Question

I found some difficulty with this exercise:

Calculate

$$\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$

with $D := \left\{(x,y)\in\mathbb{R}^{2}\mid \dfrac{x^2}{2}+y^2\leq1\right\}$

I use change of Variables in Polar Coordinates, but the integral become so hard to calculate.

I think maybe we change variables $u = x^2 + y^2$, the integral will be easier, but I can't find $v(x,y$) to have the Jacobi easy to calculate.

Answer 1

Take $$I=\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$ Consider the change of variables $(x,y)\to(r\cos\theta,r\sin\theta)$.

Let $\partial D=\{(x,y)\in\mathbb{R}^2:\frac{x^2}2+y^2=1\}$. Then for all $(x,y)=(r\cos\theta,r\sin\theta)\in\partial D$, $r^2(1-\frac12\cos^2\theta)=1$. Hence, $$I=\int_0^{2\pi}\left(\int_0^{R(\theta)}\frac{r^2}{\sqrt{4-r^4}}rdr\right)d\theta$$ where $R(\theta)=\frac1{\sqrt{1-\frac12\cos^2\theta}}$. Now,

$$I=\int_0^{2\pi}\frac12\left(2-\sqrt{4-[R(\theta)]^2}\right)d\theta$$ $$\implies I=2\pi-\frac12\int_0^{2\pi}\sqrt{4-\left(1-\frac12\cos^2\theta\right)^{-2}}d\theta$$

Answer 2

Here is a possible way to evaluate explicitly.

We use the change of variables $x=r\sqrt 2\; \cos t$, $y=r\sin t$. Then we have formally $$ \begin{aligned} dx &= \sqrt 2\; \cos t\; dr - r\sqrt 2\; \sin t\; dt\ ,\\ dy &= \sin t\; dr + r\; \cos t\; dt\ ,\\ dx\wedge dy &= dr\wedge dt\cdot \sqrt 2\; \cos t\cdot r\; \cos t - dt\wedge dr\cdot r\sqrt 2\; \sin t \cdot \sin t \\ &= dr\wedge dt\cdot r\sqrt 2\cdot ( \cos^2 t +\sin^2 t) \\ &= dr\wedge dt\cdot r\sqrt 2\ .\\[2mm] x^2 + y^2 &=2r^2\cos ^2t + r^2\sin^2 t\\ &=r^2(1+\cos^2 t)\ . \end{aligned} $$ So the given integral $I$ can be computed as follows: $$ \begin{aligned} I &=\iint_D \frac{x^2+y^2}{\sqrt {4-(x^2+y^2)^2}}\; dx\; dy \\ &=4 \iint_{substack{(x,y)\in D\\x,y\ge 0}} \frac{x^2+y^2}{\sqrt {4-(x^2+y^2)^2}}\; dx\; dy \\ &= 4 \iint_{\substack{0\le r\le 1\\0\le t\le \pi/2}} \frac{r^2(1+\cos^2 t)}{\sqrt{4-r^4(1+\cos^2 t)^2}} \;r\sqrt 2\; dr\; dt \\ &\qquad\text{ ... now use $s=r^4$, $ds=4r^3\; dr$} \\ &= \sqrt 2 \iint_{\substack{0\le s\le 1\\0\le t\le \pi/2}} \frac {\color{blue}{1+\cos^2 t}} {\sqrt{4-s\color{red}{(1+\cos^2 t)^2}}} \;ds\; dt \\ &\qquad\text{ ... now use $\int_0^1\frac{ds}{\sqrt{4-s\color{red}{a}}} =\frac 1{\color{red}{a}}(4-2\sqrt {4-\color{red}{a}})$} \\ &= \sqrt 2 \int_0^{\pi/2} \frac{\color{blue}{1+\cos^2 t}}{\color{red}{(1+\cos^2 t)^2}} \left(4-2\sqrt{4-\color{red}{(1+\cos^2 t)^2}}\right) \; dt \\ &= 2\sqrt 2 \int_0^{\pi/2} \frac1{1+\cos^2 t} \left(2-\sqrt{4-(1+\cos^2 t)^2}\right) \; dt \\ &= 2\pi - 2\sqrt 2 \int_0^{\pi/2} \frac1{1+\cos^2 t} \sqrt{2^2-(1+\cos^2 t)^2} \; dt \\ &= 2\pi - 2\sqrt 2 \int_0^{\pi/2} \frac1{1+\cos^2 t} \sqrt{(3+\cos^2 t)(1-\cos^2 t)} \; dt \\ &= 2\pi - 2\sqrt 2 \int_0^{\pi/2} \frac1{1+\cos^2 t}\cdot \sqrt{3+\cos^2 t} \; \sin t\; dt \\ &\qquad\text{ ... now use $u=\cos t$} \\ &= 2\pi - 2\sqrt 2 \int_0^1 \frac1{1+u^2}\cdot \sqrt{3+u^2} \; du \\ &= \color{forestgreen}{ 2\pi - \sqrt 2\log 3 - 4\arctan\frac 1{\sqrt 2}}\ . \end{aligned} $$ $\square$

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Numerical check:

sage: 4 * numerical_integral( lambda x:
....:        numerical_integral( lambda y :
....:            (x^2 + y^2) / sqrt(4 - (x^2 + y^2)^2),
....:                            (0, sqrt(1-x^2/2)) )[0],
....:                         (0, sqrt(2)) )[0]
2.2675940740738505
sage: ( 2*pi - sqrt(2)*log(3) - 4*atan(1/sqrt(2)) ).n()
2.26759407407385