How to show that $\frac{\pi}{3}\le \iint_D \left(x^2+(y-2)^2\right)^{-1/2}\,dx\,dy\le \pi$ where $D$ is the unit disc.

Question

How to show that $$\frac{\pi}{3}\le \iint_D \frac{dx\ dy}{\sqrt{x^2+(y-2)^2}}\le \pi$$ where $D$ is the unit disc centered at the origin?

I was trying to integrate it using polar coordinates but got stuck as using polar coordinates I get $$\int_{\theta=0}^{2\pi}\int_{r=0}^1 \frac{r dr d\theta}{\sqrt{r^2-2r \sin\theta+4}}$$

I have no idea on how to solve it further.

Also, I think we don't need to evaluate the integral and do some computations to show that it lies between $\pi/3$ and $\pi$.

Can someone please help me out? Thanks

Answer 1

Given $r\in(0,1)$ we have that $$\int_{0}^{2\pi}\frac{d\theta}{\sqrt{r^2-2r\sin\theta+4}}=\int_{-\pi}^{\pi}\frac{d\theta}{\sqrt{(1-r)^2 +3+2r(1-\cos\theta)}}$$ equals $$4\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{(1-r)^2 +3+4r\sin^2\theta}}.$$ Since $f(x)=\frac{1}{\sqrt{x^2+1}}$ is a concave function in a neighbourhood of the origin and the average value of $\sin^2\theta$ over $(0,\pi/2)$ is $\frac{1}{2}$, by Jensen's inequality the previous integral is $\geq \frac{2\pi}{\sqrt{(1-r)^2+3+2r}}$ and the wanted integral is $\geq 2\pi(\sqrt{5}-2)=\frac{\pi}{1+\frac{\sqrt{5}}{2}}>\frac{\pi}{3}$. On the other hand the previous integral is $\leq \frac{2\pi}{\sqrt{(1-r)^2+3}}$, so the wanted integral is $\leq \pi\left(-4+2\sqrt{3}+\log 3\right)<\pi$.

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The wanted integral is actually

$$ \int_{0}^{1}\frac{4r}{\sqrt{4+2r+r^2}}\,K\left(\frac{4r}{4+2r+r^2}\right)\,dr $$ according to Mathematica's notation for the complete elliptic integral of the first kind (the parameter is the elliptic modulus). Since $K$ is related to the $\text{AGM}$ mean which is rapidly convergent, the numerical evaluation of the original integral is quite simple.

Answer 2

We have $D=\{(x,y)\ |\ x^2+y^2\leq 1\}$. Clearly $x^2+(y-2)^2 \geq 1$. So we get: \begin{align} \iint_D \frac{1}{\sqrt[]{x^2+(y-2)^2}} dxdy \leq \iint_D dx dy = \pi \end{align} For the other inequality we note that the function $f(x,y)=x^2 + (y-2)^2$ can only have its maximum on the boundary of $D$ (why?). Let $\phi(x,y)=x^2+y^2-1$, so that $D=\{(x,y)\ |\ \phi(x,y)=0\}$. By Lagrange Multiplier method we must solve: \begin{align} \begin{cases} \nabla f(x,y) + \lambda \nabla \phi(x,y) = 0 \\ \phi(x,y)=0\\ \nabla \phi(x,y) \neq 0 \end{cases} \end{align} for some $\lambda\in\mathbb{R}$. Solving this is simple and yields $(x,y)=(0,-1)$ is the maximum of $f(x,y)$ on $D$. So $f(x,y)=x^2+(y-1)^2 \leq f(0,-1) = 9$. We get finally: \begin{align} \iint_D \frac{1}{\sqrt[]{x^2+(y-2)^2}} dxdy \geq \iint_D \frac{1}{\sqrt[]{9}}dx dy = \frac{\pi}{3} \end{align} So:

\begin{align} \frac{\pi}{3} \leq \iint_D \frac{1}{\sqrt[]{x^2+(y-2)^2}} dxdy \leq \pi \end{align}