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$$\leq$$f(x,y)$$\leq$$8

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.