1. ## integral

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?

2. What's the original question?

3. The question is to find the area of this surface (in the pic).
What it means, is to find the surface area of the cylinder between z=0 and z=sqrt(x^2+y^2).

4. (It is assume that $a>0$)

Let $A$ be the disk $(x-a)^2+y^2 = a^2$.
Then what you are looking for is $\iint_A \sqrt{1+(\partial_1 f)^2+(\partial_1 f)^2}$.
Where $f(x,y) = \sqrt{x^2+y^2}$.

This means, $\partial_1 f(x,y) = - \frac{x}{\sqrt{x^2+y^2}}$ and $\partial_2 f(x,y) = -\frac{y}{\sqrt{x^2+y^2}}$

Thus, we end up with $\iint_A \sqrt{2} = \sqrt{2} \cdot \text{area}(A) = \sqrt{2}\pi a^2$