# Surface integral

1. May 13, 2010

### bosox284

The problem statement, all variables and given/known data
Evaluate the surface integral, the double integral of zdS if the region is the patch of surface defined by x^2 + y^2 - z^2 = -1 in the first octant with z less than or equal to 4.

The attempt at a solution
I really don't know where to begin. I believe the equation is of a hyperboloid of two sheets. So I want to say I have to parametrize the equation with x = bxcos(u)sinh(v), y = bysin(u)sinh(v), z = bzcosh(v), and then take the partial derivatives of u and v, and take the cross product of the two. Could someone please point me in the right direction? Or instruct me where to go from there?

2. May 13, 2010

### gabbagabbahey

That sounds like a good plan, although I see no need for the $b$'s...why not show us how far you get with that method?

3. May 13, 2010

### bosox284

Well the b's would be for if it were like (x^2)/4, where b would be 2. So in this case, the value of b is one and I could have gone without them. The only problem I see myself running into from here, would probably be finding out the values of u and v for when I set up the integral.

4. May 13, 2010

### gabbagabbahey

That shouldn't be too difficult...$x$,$y$, and $z$ all have to be positive in the first octant and $z\leq 4$....find the range on $v$ first using your limits on $z$ and then see what range of $u$-values $x$ and $y$ are positive for.