# Surface Integral and the unit circle

learning surface integrals now.
I have a surface which is composed of three surfaces, one of them is the unit circle on the x-y plane. How would I find the surface integral of this first surface (this unit circle)? Isn't it just the area of a circle? If so, how would I do this the "surface integral" method? As in, finding a parametrization for this surface and evaluating the integral?
The parametrization to me looks like $$r(u,v) = \cos u i + \sin u j + 0k$$ which, when finding the cross product of the partials, obviously evaluates to zero... meaning the integral (i.e. surface area) is zero? Why is this not correct?

No, the integral depends on the function you're integrating. If the function is $1$ then the integral will be the area of the circle, but for almost any other function there is no reason for the integral to be the area of the circle. This is similar to asking in 1D integrals if computing the integral over some interval gives you its length. Well, for $f(x)=1$, $$\int_a^b f(x)dx = b-a$$ But for almost any other function this isn't correct because the function gives different weight to different parts of the interval/area. For a uniform weight of 1 the sum of the weights will be the area.
$$r(u,v) = v\cos u \hat{i} + v \sin u \hat{j} + 0 \hat{k}$$