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)=[tex]\sqrt{x^2+y^2}[/tex]
0[tex]\leq[/tex]f(x,y)[tex]\leq[/tex]8

Homework Equations

**The answer is [tex]\sqrt{2}\pi[/tex]**

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

Reduces to
[tex]\sqrt{2}[/tex]

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

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.