<s>[INST] Determine the geometric centre of a circle with a quarter missing The question
I have a circle of radius $a$, (where $a$ is a known positive constant), centred at Cartesian coordinates $(a,a)$. The bottom left quarter of the circle is missing.
Let the two-dimensional region $R$ be formed by this circle with a missing quarter. I'd like to compute the geometric centre of the region $R$ (or the centre of mass if we were to assume it has uniform density $\rho(x,y)=1)$.
My attempt
I know that for uniform density, the geometric centre of a region $R$, $(\bar x, \bar y)$ is
$$\bar x={1\over|R|}\iint_Rx\,\mathrm dA$$
$$\bar y={1\over|R|}\iint_Ry\,\mathrm dA$$
where $|R|$ is the area of the region. I got the area of the region by $|R|=\frac{3\pi a^2}{4}$.
I recognise that these integrals are better suited to be calculated in polar coordinates so I parametrise as such:
$$\begin{cases}x&=a\cos(\theta)+a\\y&=a\sin(\theta)+a\end{cases}$$
where $-\frac{\pi}{2}\leq\theta\leq\pi$.
I will then show a calculation for $\bar x$, which is erroneous (the calculation for $\bar y$ is the same):
$$\begin{align*}\bar x&={1\over |R|}\iint_R x\,\mathrm dA\\&={1\over |R|}\int_?^?\int_?^?(a\cos(\theta)+a)\,\mathrm da\,\mathrm d\theta\end{align*}$$
I think I may have made a mistake setting up the integral limits--this part confuses me:
$$\begin{align*}&={1\over |R|}\int_{-\pi\over 2}^\pi\int_0^a(a\cos(\theta)+a)\,\mathrm da\,\mathrm d\theta\\&=\frac{4}{3\pi a^2}\times\frac{3\pi a^2+2a}{4}\\&=\frac{2+3\pi}{4}\end{align*}$$
which does not even depend on $a$. I have clearly done something wrong.
 [/INST] Choose polar coordinates centered at the center of your circle so the change-of-variables looks like
\begin{cases} 
x = a + r \cos \theta, \\
y = a + r \sin \theta. 
\end{cases}
Notice that when $r=0$, we're at the center $(x, y) = (a, a)$. As you correctly observed, the region of interest is defined by the inequalities
$$
0 \leq r \leq a 
\quad \text{and} \quad 
-\tfrac{\pi}{2} \leq \theta \leq \pi. 
$$
The Jacobian determinant for the change of variables to these shifted polar coordinates are the same as for standard polar coordinates:
$$
\frac{\partial(x, y)}{\partial(r, \theta)} 
= 
\left\lvert 
\begin{matrix}
\partial x/ \partial r & \partial x/ \partial \theta \\
\partial y/ \partial r & \partial y/ \partial \theta 
\end{matrix}
\right\rvert
=
\left\lvert 
\begin{matrix}
\cos \theta & -r \sin \theta \\
\sin \theta & r \cos \theta 
\end{matrix}
\right\rvert
= r \, (\cos^2 \theta + \sin^2 \theta)
= r 
$$
Let's calculate $\bar{x}$ since, by symmetry, $\bar{y} = \bar{x}$.
\begin{align}
\bar{x} &= \frac{1}{\frac{3}{4} \pi a^2} 
\int_{-\tfrac{\pi}{2}}^{\pi} \int_0^a (a + r\cos\theta)\, r \, \mathrm{d}r \, \mathrm{d}\theta \\
&= \frac{4}{3 \pi a^2} 
\int_{-\tfrac{\pi}{2}}^{\pi} \int_0^a (ar + r^2\cos\theta) \, \mathrm{d}r \, \mathrm{d}\theta \\
&= \frac{4}{3 \pi a^2} 
\int_{-\tfrac{\pi}{2}}^{\pi} \left.\biggl( \frac{ar^2}{2} 
+ \frac{r^3\cos\theta}{3} \biggr)\right\rvert_0^a \, \mathrm{d}\theta \\
&= \frac{4}{3 \pi a^2} \cdot \frac{a^3}{6} 
\int_{-\tfrac{\pi}{2}}^{\pi} \bigl( 3 + 2\cos\theta \bigr) \, \mathrm{d}\theta \\
&= \frac{2a}{9 \pi} \Bigl. \bigl( 3\theta + 2\sin\theta \bigr) 
\Bigr\rvert_{-\tfrac{\pi}{2}}^{\pi} \\
&= \frac{2a}{9 \pi} \cdot \biggl( 3 \cdot \frac{3\pi}{2} + 2 \cdot 1 \biggr) \\
&= \frac{(9\pi + 4) \, a}{9 \pi}
\end{align}
Here's an interactive picture. You can drag the center of the circle and see the center of mass.
</s>