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.
Best Answer
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.