# integral

• August 11th 2008, 10:35 AM
asi123
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?
• August 11th 2008, 11:02 AM
wingless
What's the original question?
• August 11th 2008, 11:08 AM
asi123
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).
• August 11th 2008, 12:42 PM
ThePerfectHacker
(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 $\iint_A \sqrt{2} = \sqrt{2} \cdot \text{area}(A) = \sqrt{2}\pi a^2$