Evaluate the integral $\int_{C} z ds$ where C is the intersection of $x^{2}+y^{2}=4$ and $z=0$ (oriented clockwise as viewed from above).
My interpretation of this problem yields the following conflict:
First: parameterization of the circle $x^{2}+y^{2}=4$ yields $x=2\cos{t}$;$y=2\sin{t}$ where $z=x^{2}+y^{2} -4$ given that $z=0$ this then yields $z=4\cos^{2}{t}+4\sin^{2}{t} -4 =0$ Which is validated by the answer's prompt.
Evaluating this integral one finds
$$\int_{C} zds= \int_{t=0}^{t=2\pi} dt =2\pi $$
My intuition conflicts with this result because I think that we're trying to derive, via calculus, the circumference of the circle $x^{2}+y^{2}=4$. Which give $C=\pi d \rightarrow C=4\pi$.
The only place I may be missing something is the fact that we're integrating over z as our function. I would greatly appreciate any help in reconciling these two ideas.
This is how to calculate the circumference. It is not exactly what you're asking, but it's a bit too long for a comment, so I'm doing it in an answer.
The length of a curve $C$ is calculated by integrating $\int_C ds$ (if you dislike the empty integrand, we can write $\int_C 1\,ds$ which is equivalent). Substituting a (differentiable) parametrisation $C:t \mapsto(x(t), y(t))$ for $t\in [a, b]$, we get that this is equal to $$ \int_{t = a}^{t =b}\sqrt{x'(t)^2 + y'(t)^2}\:dt $$ Interpreting this as a Riemann sum, and dividing $[a, b]$ into lots of tiny pieces of size $\Delta t$, it tells us that the length of the parametrised curve $C$ is equal to the sum of the lengths of all the little parts of the parametrized curve. Each of those, in turn, is equal to $\Delta t$ times the speed of the parametrisation at that $t$-point, which is $\sqrt{x'(t)^2 + y'(t)^2}$.
In this case, we can calculate the integrand. We have $x'(t) = -2\sin t$ and $y'(t) = 2\cos t$, so $$ \sqrt{4(-\sin t)^2 + 4(\cos t)^2} = 2\sqrt{\sin^2t+\cos^2t} = 2 $$ so the entire integral (and thus the circumference) evaluates to $$ \int_C ds = \int_{t = 0}^{t =2\pi}\sqrt{x'(t)^2 + y'(t)^2}\:dt = \int_{t = 0}^{t = 2\pi}2\,dt = 4\pi $$