Stoke's theorem, why is this the surface?

[flyingpig]

Homework Statement

Evaluate $$\int \mathbf{F} \cdot d\mathbf{S}$$ where $$\mathbf{F} = <-y^2,x,z^2>$$ and C is the curve of intersection of the plane y + z = 2 and the cylinder $$x^2 + y^2 = 1$$ (Orient C to be CCW when viewed from above)

Solution

curlF = <0,0,1+2y>

z = 2 - y

$$\int \mathbf{F} \cdot d\mathbf{S} = \iint_S curl \mathbf{F} \cdot d\mathbf{S} = \int_{0}^{2\pi} \int_{0}^{1} (1 + 2r\; \sin\theta) r dr d\theta = \pi$$

The Attempt at a Solution

I did all of that, except I don't understand why the surface is chosen to be the plane. I thought the surface I am suppose to integrate is the cylinder

So parametrizing

$$\mathbf{r}(u,v) = <cos(u), sin(u), v>$$

$$\mathbf{r_u} \times \mathbf{r_v} = <cos(u),sin(u),0>$$

$$\iint_S curl\mathbf{F} \cdot d\mathbf{S} = \iint_S 0 dS = 0$$

The 0 is wrong i know...

 

[lanedance]

so you're trying to compute the line integral of C, the intersection of a plane and cylinder, which is an ellipsoid. You do this by using stokes theorem to convert to a surface integral, the surface can be anyone that has boundary C, so the plane is the logical choice as it is the simplest to parameterise