Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 3: Many Variables; Section 2: Integration; page 3

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:

The Jacobian is

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

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

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

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:

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

Parts 3-5

Solution of question 2.

3. According to (3.2.3.34) and with the help of parts 1 and 2 we get
   
   $$A=\underset{{\phi }_{\min }}{\overset{{\phi }_{\max }}{\int }}\text{d}\phi \underset{a}{\overset{2a\cos \phi }{\int }}r\text{d}r$$


4. The limits for the integration over  φ  are obtained from the intersection of the circles:
   
   $$a=2a\cos \phi$$ yielding $$\begin{array}{l}\cos \phi =\frac{1}{2}\\ \phi =\pm \frac{\pi }{3}\end{array}\}$$
5. The integral becomes
   
   $$\begin{array}{l}A=\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\text{d}\phi \underset{a}{\overset{2a\cos \phi }{\int }}r\text{d}r=\frac{a^2}{2}\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\text{d}\phi \left(4{\cos }^2\phi -1\right)=\\ =\frac{a^2}{2}\left(2\sin \phi \cos \phi +2\phi -\phi \right)\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{|}}=\\ =\frac{a^2}{2}2\left(\frac{2\sqrt{3}}{4}+\frac{\pi }{3}\right)=a^2\left(\frac{\sqrt{3}}{2}+\frac{\pi }{3}\right)\end{array}$$

Parts 6-8

Solution of question 2.

6. Since the body is uniform, the centre of mass is equivalent to the centroid
   
   $$x_{cm}=\langle x\rangle$$ By definition $$\langle x\rangle A=\int x\text{d}x\int \text{d}y$$
   which after the transformation to polar coordinates, and by applying the limits of integration from part 5, becomes
   
   $$\langle x\rangle A=\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\text{d}\phi \cos \phi \underset{a}{\overset{2\text{a}\text{cos}\phi }{\int }}{r}^2\text{d}r$$
7. yielding $$\begin{array}{l}\langle x\rangle A=\frac{a^3}{3}\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\text{d}\phi \cos \phi \left(8{\cos }^3\phi -1\right)=\\ =\frac{a^3}{3}\left[8\left(\frac{3\phi }{8}+\frac{\sin \left(2\phi \right)}{4}+\frac{\sin \left(4\phi \right)}{32}\right)-\sin \phi \right]\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{|}}=\\ =\frac{a^3}{3}\left[3\phi +2\sin \left(2\phi \right)+\frac{\sin \left(4\phi \right)}{4}-\sin \phi \right]\underset{-\frac{\pi }{3}}{\overset{\frac{\pi }{3}}{|}}=\\ =\frac{2a^3}{3}\left(\frac{3\pi }{3}+\frac{2\sqrt{3}}{2}-\frac{\sqrt{3}}{8}-\frac{\sqrt{3}}{2}\right)=\\ =a^3\left[\frac{2\pi }{3}+\frac{\sqrt{3}}{3}\left(2-\frac{1}{4}-1\right)\right]=a^3\left(\frac{2\pi }{3}+\frac{\sqrt{3}}{4}\right)\end{array}$$
   where a table of integrals was used, and the value of the angle was substituted:
   
   $$\sin \phi =\frac{\sqrt{3}}{2}\sin 2\phi =\frac{\sqrt{3}}{2}\sin 4\phi =-\frac{\sqrt{3}}{2}$$
8. Finally we get $$x_{cm}=\frac{\langle x\rangle A}{A}=a\frac{\frac{2\pi }{3}+\frac{\sqrt{3}}{4}}{\frac{\pi }{3}+\frac{\sqrt{3}}{2}}=\frac{8\pi +3\sqrt{3}}{2\left(2\pi +3\sqrt{3}\right)}\approx 1.321a$$

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.