How Is the Surface Area of z=sqrt(x^2+y^2) Calculated?

[filter54321]

Homework Statement

What's the surface area of the following 3D curve over the restricted range:
z=f(x,y)=\sqrt{x^2+y^2}
0\leqf(x,y)\leq8

Homework Equations

**The answer is \sqrt{2}\pi**

The surface area equation (with partials)
\sqrt{1+(Fx)^2+(Fy)^2}

Reduces to
\sqrt{2}

So, for an as yet unknown integration range, we have
\int\int\sqrt{2}dydx

The Attempt at a Solution

Since the Z is restricted to [0,8] it would seem x and y should both be limited to [-8,8] but that integration range doesn't compute the the correct answer (listed above).

What's the range of integral for both dy and dx?

 

[Dick]

sqrt(x^2+y^2)=8 is a circle of radius 8, isn't it? What does that tell you about the domain? But I don't see how you are going to get sqrt(2)*pi out of that.