Hey guys.
I'm in the middle of this crazy exercise and I got stuck in this ugly integral in the bottom of the page, any ideas on how to solve it?
(It is assume that $\displaystyle a>0$)
Let $\displaystyle A$ be the disk $\displaystyle (x-a)^2+y^2 = a^2$.
Then what you are looking for is $\displaystyle \iint_A \sqrt{1+(\partial_1 f)^2+(\partial_1 f)^2}$.
Where $\displaystyle f(x,y) = \sqrt{x^2+y^2}$.
This means, $\displaystyle \partial_1 f(x,y) = - \frac{x}{\sqrt{x^2+y^2}}$ and $\displaystyle \partial_2 f(x,y) = -\frac{y}{\sqrt{x^2+y^2}}$
Thus, we end up with $\displaystyle \iint_A \sqrt{2} = \sqrt{2} \cdot \text{area}(A) = \sqrt{2}\pi a^2 $