I have to prove that the maximum/minimum value of $$ I(D)=\iint_{D}\frac{dx\, dy}{\sqrt{x^2+y^2}} $$ where $D$ is a unit disk in $\mathbb{R}^2$ enclosing the origin, are respectively $\pi$ and $\frac{\pi}{3}$.
I am not able begin it. I need some hint for approaching the problem.
$$\mathscr{I}_{\space\text{n}}\left(\text{a},\text{b},\text{c},\text{d}\right):=\int_\text{a}^\text{b}\int_\text{c}^\text{d}\frac{1}{\sqrt{x^2\cdot\left(\text{y}-\text{n}\right)^2}}\space\text{d}\text{y}\space\text{d}x=$$ $$\int_\text{a}^\text{b}\int_\text{c}^\text{d}\frac{1}{x\cdot\left(\text{y}-\text{n}\right)}\space\text{d}\text{y}\space\text{d}x\tag1$$
Now, substitute $\text{u}:=\text{y}-\text{n}$:
$$\mathscr{I}_{\space\text{n}}\left(\text{a},\text{b},\text{c},\text{d}\right)=\int_\text{a}^\text{b}\left\{\frac{1}{x}\int_{\text{c}-\text{n}}^{\text{d}-\text{n}}\frac{1}{\text{u}}\space\text{d}\text{u}\right\}\space\text{d}x=\int_\text{a}^\text{b}\left\{\frac{1}{x}\cdot\left[\ln\left|\text{u}\right|\right]_{\text{c}-\text{n}}^{\text{d}-\text{n}}\right\}\space\text{d}x=$$ $$\int_\text{a}^\text{b}\left\{\frac{1}{x}\cdot\left(\ln\left|\text{d}-\text{n}\right|-\ln\left|\text{c}-\text{n}\right|\right)\right\}\space\text{d}x=\int_\text{a}^\text{b}\left\{\frac{1}{x}\cdot\left(\ln\left|\frac{\text{d}-\text{n}}{\text{c}-\text{n}}\right|\right)\right\}\space\text{d}x=$$ $$\ln\left|\frac{\text{d}-\text{n}}{\text{c}-\text{n}}\right|\int_\text{a}^\text{b}\frac{1}{x}\space\text{d}x=\ln\left|\frac{\text{d}-\text{n}}{\text{c}-\text{n}}\right|\cdot\ln\left|\frac{\text{b}}{\text{a}}\right|\tag2$$