# 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.

• Note that $C=\{(x,y)\in \mathbb R^2\colon x^2+y^2=16\land x\ge 0\}=\left\{(4\cos(\theta), 4\sin(\theta))\colon \theta \in \left[-\frac \pi 2, \frac \pi 2\right]\right\}$. Commented Nov 13, 2014 at 17:47
• @GitGud If you'd like to post your comment as an answer, I think that hint was sufficient to make the problem understandable. Commented Nov 13, 2014 at 17:58
• I don't want to because if I were to answer, I'd like to push your idea through (which can work), but I can't answer at the moment. I'll answer in a few hours if no one does in the mean time. Commented Nov 13, 2014 at 17:59
• Made your idea work. In my opinion it's not harder than what I suggested at first. Commented Nov 14, 2014 at 2:26

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.

Take $x = 4cos(t)$ , y =$4sin(t)$ your bounds for t are from -pi/2 to +pi/2 .