]> Exercise 4

# Two Dimensional Transformation, Exercise 4

On the (x,y) plane there are two circles with the same radius  a : the first is centred at the origin and the second - at the point (x,y)=(a,0). The second circle, excluding the overlapping part with the first, is a solid uniform planar body with density  σ .

Use polar coordinates to answer the following questions.

1. Express the two circles in polar coordinates!
2. Calculate the area  A  of the planar body!
3. What is the x coordinate of its centre of mass?

## Reminder

..... relation of polar with the Cartesian coordinates is:

 $\begin{array}{l}\left\{\begin{array}{l}x=r\mathrm{cos}\phi \\ y=r\mathrm{sin}\phi \end{array}\\ \left\{\begin{array}{l}r\ge 0\\ \Phi \le \phi <\Phi +2\pi \\ \text{e}\text{.g}\text{.}\text{ }\Phi =0\end{array}\end{array}\right\}$ (3.2.3.19)

The Jacobian is

 $\frac{D\left(x,y\right)}{D\left(r,\phi \right)}=\frac{\partial x}{\partial r}\frac{\partial y}{\partial \phi }-\frac{\partial x}{\partial \phi }\frac{\partial y}{\partial r}=r\left({\mathrm{cos}}^{2}\phi +{\mathrm{sin}}^{2}\phi \right)=r$ (3.2.3.20)

Therefore an area element of polar coordinates, in units of Cartesian coordinates, is given by:

 $\Delta x\Delta y=r\text{​}\Delta r\text{​}\Delta \phi$ (3.2.3.21)

......................

The integration of a function f(r,φ) over a region with a boundary r(φ), including the origin of the coordinates inside or on the boundary, is

 $\underset{{\phi }_{\mathrm{min}}}{\overset{{\phi }_{\mathrm{max}}}{\int }}\text{d}\phi \underset{0}{\overset{r\left(\phi \right)}{\int }}f\left(r,\phi \right)\text{​}r\text{​}\text{d}r$ (3.2.3.23)

The vanishing lower limit of the integral with respect to r, is due to the inclusion of the origin in the region of integration.

..................

For a region of integration, which does not include the origin, the limits of the variable for the first integration, should be functions of the second variable, similarly to the Cartesian case:

 $\begin{array}{l}\underset{{\phi }_{\mathrm{min}}}{\overset{{\phi }_{\mathrm{max}}}{\int }}\text{d}\phi \underset{{r}_{\mathrm{min}}\left(\phi \right)}{\overset{{r}_{\mathrm{max}}\left(\phi \right)}{\int }}f\left(r,\phi \right)\text{ }r\text{d}r\\ \underset{{r}_{\mathrm{min}}}{\overset{{r}_{\mathrm{max}}}{\int }}r\text{d}r\underset{{\phi }_{\mathrm{min}}\left(r\right)}{\overset{{\phi }_{\mathrm{max}}\left(r\right)}{\int }}f\left(r,\phi \right)\text{ }\text{d}\phi \end{array}\right\}\text{\hspace{0.17em}}\text{or}$ (3.2.3.34)

It is illustrated in Fig. Region excluding the origin.

## Parts 1-2

Solution of question 1.

1. The representation of the first circle in Cartesian coordinates is

${x}^{2}+{y}^{2}={a}^{2}$
The application of the transformation (3.2.3.19) gives

${x}^{2}+{y}^{2}={r}^{2}\left({\mathrm{cos}}^{2}\phi +{\mathrm{sin}}^{2}\phi \right)={r}^{2}={a}^{2}$ yielding $r=a$

2. The representation of the second circle in Cartesian coordinates is

${\left(x-a\right)}^{2}+{y}^{2}={a}^{2}$
The application of the transformation (3.2.3.19) gives

${\left(r\mathrm{cos}\phi -a\right)}^{2}+{r}^{2}{\mathrm{sin}}^{2}\phi ={r}^{2}-2ar\mathrm{cos}\phi +{a}^{2}={a}^{2}$ yielding $r=2a\text{ }\mathrm{cos}\phi$

## Parts 3-5

Solution of question 2.

1. According to (3.2.3.34) and with the help of parts 1 and 2 we get

$A=\underset{{\phi }_{\mathrm{min}}}{\overset{{\phi }_{\mathrm{max}}}{\int }}\text{d}\phi \underset{a}{\overset{2a\text{ }\mathrm{cos}\phi }{\int }}r\text{d}r$

2. The limits for the integration over  φ  are obtained from the intersection of the circles:

$a=2a\text{ }\mathrm{cos}\phi$ yielding $\text{ }\begin{array}{l}\text{ }\mathrm{cos}\phi =\frac{1}{2}\\ \phi =±\frac{\pi }{3}\end{array}\right\}$
3. The integral becomes

$\begin{array}{l}A=\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\text{d}\phi \underset{a}{\overset{2a\text{ }\mathrm{cos}\phi }{\int }}r\text{d}r=\frac{{a}^{2}}{2}\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\text{d}\phi \left(4{\mathrm{cos}}^{2}\phi -1\right)=\\ =\frac{{a}^{2}}{2}\left(2\mathrm{sin}\phi \mathrm{cos}\phi +2\phi -\phi \right)\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{|}}=\\ =\frac{{a}^{2}}{2}2\left(\frac{2\sqrt{3}}{4}+\frac{\pi }{3}\right)={a}^{2}\left(\frac{\sqrt{3}}{2}+\frac{\pi }{3}\right)\end{array}$

## Parts 6-8

Solution of question 2.

1. Since the body is uniform, the centre of mass is equivalent to the centroid

${x}_{cm}=〈x〉$ By definition $〈x〉A=\int x\text{d}x\int \text{d}y$
which after the transformation to polar coordinates, and by applying the limits of integration from part 5, becomes

$〈x〉A=\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\text{d}\phi \text{ }\mathrm{cos}\phi \underset{a}{\overset{2\text{a}\text{\hspace{0.17em}}\text{cos}\phi }{\int }}{r}^{2}\text{d}r$
2. yielding $\begin{array}{l}〈x〉A=\frac{{a}^{3}}{3}\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\text{d}\phi \text{ }\mathrm{cos}\phi \left(8{\mathrm{cos}}^{3}\phi -1\right)=\\ =\frac{{a}^{3}}{3}\left[8\left(\frac{3\phi }{8}+\frac{\mathrm{sin}\left(2\phi \right)}{4}+\frac{\mathrm{sin}\left(4\phi \right)}{32}\right)-\mathrm{sin}\phi \right]\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{|}}=\\ =\frac{{a}^{3}}{3}\left[3\phi +2\mathrm{sin}\left(2\phi \right)+\frac{\mathrm{sin}\left(4\phi \right)}{4}-\mathrm{sin}\phi \right]\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{|}}=\\ =\frac{2{a}^{3}}{3}\left(\frac{3\pi }{3}+\frac{2\sqrt{3}}{2}-\frac{\sqrt{3}}{8}-\frac{\sqrt{3}}{2}\right)=\\ ={a}^{3}\left[\frac{2\pi }{3}+\frac{\sqrt{3}}{3}\left(2-\frac{1}{4}-1\right)\right]={a}^{3}\left(\frac{2\pi }{3}+\frac{\sqrt{3}}{4}\right)\end{array}$
where a table of integrals was used, and the value of the angle was substituted:

$\mathrm{sin}\phi =\frac{\sqrt{3}}{2}\text{ }\mathrm{sin}2\phi =\frac{\sqrt{3}}{2}\text{ }\mathrm{sin}4\phi =-\frac{\sqrt{3}}{2}$
3. Finally we get ${x}_{cm}=\frac{〈x〉A}{A}=a\frac{\frac{2\pi }{3}+\frac{\sqrt{3}}{4}}{\frac{\pi }{3}+\frac{\sqrt{3}}{2}}=\frac{8\pi +3\sqrt{3}}{2\left(2\pi +3\sqrt{3}\right)}\approx 1.321a$

## Score

By parts.
Parts 1,2,3,4,6,8 are worth 1 point each.
Parts 5,7 are worth 2 points each.

By questions.
Question 1 is worth 2 points.
Questions 2,3 are worth 4 points each.