# Surface Integrals problem

1. Apr 23, 2007

### benji55545

1. The problem statement, all variables and given/known data

Evaluate ∫∫S √(1 + x^2 + y^2) dS where S is the helicoid: r(u, v) = ucos(v)i + usin(v)j + vk, with 0 ≤ u ≤ 4, 0 ≤ v ≤ 3π

3. The attempt at a solution

What I tried to do was say x=ucos(v) and y=usin(v), then I plugged those into the sqrt(1+x^2+y^2) eq, which I ended up simplifying to sqrt(1+u) somehow. I then took the cross product of the surface eq differentiated (respect to u) with eq (respect to v). I found the length of that, which I somehow got to be sqrt(2).

I then said that the intergal was ∫∫sqrt(1+u)*sqrt(2)du*dv
where the limits of integration are the limits of u and v given above.
Well this is wrong. Hah!

Any help would be appreciated.

2. Apr 24, 2007

### HallsofIvy

Staff Emeritus
I'm not happy with the "somehow"!! That's where you went wrong. The cross product you speak of (also called the 'fundamental vector product') is $sin(v)\vec{i}- cos(v)\vec{j}+ u\vec{k}$
(did you forget the "u" multiplying -cos(v) and sin(v) in the derivative with respect to v?) and that has length $\sqrt{1+ u^2}$. Thus your integral is
$$\int_{u= 0}^4\int_{v=0}^{2\pi}(1+ u^2)dudv$$

Last edited: Apr 24, 2007
3. Apr 24, 2007