Evaluate $\int_C (x + y)\text ds$, where $C$ is the circle centred at $(1/2, 0)$ with radius $1/2$.
parametrise
$$x=\frac12\cos(t) \tag1$$
$$y=\frac12\sin(t)\tag2$$
$2\pi\geq t\geq0$
$$ds=\sqrt{dx^2+dy^2}$$ $$=\sqrt{\left(\frac12\right)^2-\sin^2(t)+\left(\frac12\right)^2\cos^2(t)}$$
$$=\sqrt{-(1)^2\left(\frac12\right)^2\sin^2(t)+\left(\frac12\right)^2\cos^2(t)}$$
$$=\sqrt{-(1)^2\left(\frac12\right)^2\left(\sin^2(t)+\cos^2(t)\right)}$$
$$ds=\frac12$$
$$\int \left[\frac12\cos(t) + \frac12\sin(t)\right]\cdot\frac12 dt$$ , where $2\pi\geq t\geq0$
I evaluated this integral and got 0. I don't think this is correct though. Does the origin of this circle starting at $(1/2,0)$ change the integral?
There are two reasonable parameterizations as I see it. The first is as all have recommended, simply translate the origin of polar coordinates to the center of the circle and then $x=\frac12+\frac12\cos t$ and $y=\frac12\sin t$, so $$ds=\sqrt{(dx)^2+(dy)^2}=\sqrt{\left(-\frac12\sin t\right)^2+\left(\frac12\cos t\right)^2}\,dt=\frac12dt$$ So the integral is $$I=\int_0^{2\pi}\left[\frac12+\frac12\cos t+\frac12\sin t\right]\cdot\frac12dt=\frac14(2\pi)(1+0+0)=\frac{\pi}2$$ Where we have used the average value of $1$ of $1$ and the average values of $\cos t$ and $\sin t$ of $0$ over the interval of length $2\pi$.
The other possibility is using the 'polar coordinate' version of the circle $r=\cos t$, $x=r\cos t=\cos^2t$, and $y=r\sin t=\sin t\cos t$, so $$ds=\sqrt{(dx)^2+(dy)^2}=\sqrt{\left(-2\sin t\cos t\right)^2+\left(\cos^2t-\sin^2t\right)^2}=\sqrt{\left(\cos^2t+\sin^2t\right)^2}=dt$$ Then $$I=\int_{-\frac{\pi}2}^{\frac{\pi}2}(\cos^2t+\sin t\cos t)dt=\pi(\frac12+0)=\frac{\pi}2$$ Where we have used the average value of $\cos^2t$ of $\frac12$ and of $\sin t\cos t=\frac12\sin2t$ of $0$ over the interval of length $\pi$.
You knew the answer up front because it should be the average value of $x$ plus the average value of $y$ over the curve times the arc length of the curve, so $$I=\left(\frac12+0\right)(2\pi)(\frac12)=\frac{\pi}2$$ Where we have used the mensuration formula $s=2\pi r$ for the perimeter of a circle of radius $r$.