Line Integral Over the Right Half of a Circle $\int_Cxy^4\,\mathrm{d}s$, $C$ is the right half of the circle $x^2+y^2=16$
I solved for $x$ and got $x=\sqrt{16-y^2}$ but I'm pretty stumped about where I should go. I know I need to get $x$ and $y$ as functions of $t$ but how I do this is a bit of a mystery to me especially since $y=\sqrt{16-x^2}$ and $y=-\sqrt{16-x^2}$ would be needed for the right side of the circle.
 A: 
I know I need to get $x$ and $y$ as functions of $t$

This not how I think about it at all.
One needs to find a parametrization for $C$, i.e. one needs to find a differentiable function $\varphi \colon [a,b]\to \mathbb R^2$ (with $a,b$ possibly infinite) such that $\varphi\left[[a,b]\right]=C$. There are infinite of these.
The most natural for most people is to note that $C=\{(x,y)\in \mathbb R^2\colon x^2+y^2=16\land x\ge 0\}=\left\{(4\cos(\theta), 4\sin(\theta))\colon \theta \in \left[-\frac \pi 2, \frac \pi 2\right]\right\}$ and the last form of the set suggests taking $\varphi\colon \left[-\frac \pi 2, \frac \pi 2\right]\to \mathbb R^2, \theta\mapsto (4\cos(\theta), 4\sin(\theta))$.
One then gets
$$
\begin{align}
\int _Cxy^4ds&=\int \limits_{- \pi /2}^{\pi /2}4\cos(\theta)\left(4\sin(\theta)\right)^4\cdot \Vert \varphi'(\theta)\Vert\,\mathrm d\theta\\
&=4^6\int \limits_{- \pi /2}^{\pi /2}\sin'(\theta)(\sin(\theta))^4\,\mathrm d\theta\\
&\,\,\vdots\\
&=2\dfrac{4^6}5.
\end{align}
$$
But what you did also works.
Note that $C=\left\{(x,y)\in \mathbb R^2\colon x=\sqrt{16-y^2}\land y\in [-4,4]\right\}=\left\{\left(\sqrt{16-t^2}, t\right)\colon t\in [-4,4]\right\}$.
So, taking $\psi\colon [-4,4]\to \mathbb R^2, t\mapsto \left(\sqrt{16-t^2},t\right)$, it comes $\psi'\colon ]-4,4[\to \mathbb R^2, t\mapsto \left(-\dfrac{t}{\sqrt{16-t^2}},1\right)$ and $\forall t\in]-4,4[\left(\left\Vert \psi'(t)\right\Vert=\dfrac{4}{\sqrt{16-t^2}}\right)$.
Thus
$$
\begin{align}
\int _Cxy^4ds&=\int \limits_{- 4}^{4}\sqrt{16-t^2}t^4\cdot \dfrac{4}{\sqrt{16-t^2}}\,\mathrm dt\\
&=4\int \limits_{- 4}^{4}t^4\,\mathrm dt\\
&\,\,\vdots\\
&=2\dfrac{4^6}5.
\end{align}
$$

This is a bit of a mystery to me especially since $y=\sqrt{16-x^2}$ and $y=-\sqrt{16-x^2}$ would be needed for the right side of the circle.

Set $$C_{\mathbf +}=\{(x,y)\in \mathbb R^2\colon x^ 2+y^2=16\land x\ge 0\land y\ge 0\}$$ and $$C_{\mathbf -}=\{(x,y)\in \mathbb R^2\colon x^ 2+y^2=16\land x\ge 0\land y\leq 0\}.$$ 
The intersecting point will not be an issue. Note that $C=C_{\mathbf +}\cup C_{\mathbf -}$, so $\displaystyle \int _C xy^4\,\mathrm ds=\int _{C_{\mathbf +}}xy^4\,\mathrm ds+\int _{C_{\mathbf -}}xy^4\mathrm ds$.
Since $$C_{\mathbf +}=\left\{(t,\sqrt{16-t^2})\colon t\in[0,4]\right\}$$
and $$C_{\mathbf -}=\left\{(t,-\sqrt{16-t^2})\colon t\in[0,4]\right\},$$
the rest comes very similarly to what was done before.
A: Take $x = 4cos(t)$  , y =$ 4sin(t)$  your bounds for t are from -pi/2 to +pi/2 .
