]> Two dimensional transformation

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

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# Two Dimensional Transformation

## Substitution

Before discussing the change of variables in a double integral, let's return to the change of one variable, which we know as substitution. Formally the change of a variable (x) with another one (u) is

 $\begin{array}{l}x=x\left(u\right)\\ \int F\left(x\right)\text{d}x=\int F\left[x\left(u\right)\right]\frac{\text{d}x}{\text{d}u}\text{d}u\end{array}\right\}$ (3.2.3.1)

In the case of a definite integral, this expression can be seen as a sum of terms containing infinitesimal increments of  x  and  u. Each term can be expressed as

 $\begin{array}{l}F\left(x\right)\Delta x=F\left[x\left(u\right)\right]\frac{\text{d}x}{\text{d}u}\Delta u\\ \text{or}\text{ }\Delta x=\frac{\text{d}x}{\text{d}u}\Delta u\end{array}\right\}$ (3.2.3.2)

the derivative multiplying  Δu  at (3.2.3.2) can be considered as a magnifying factor in order to match the size of  Δu  with that of  Δx. It may be positive or negative. This factor also takes care of balancing the physical dimensions.

The change of two variables (with respect to differentiation) was studied in the previous section. Formally it is expressed by:

 $\begin{array}{l}x=x\left(u,v\right)\\ y=y\left(u,v\right)\end{array}\right\}$ (3.2.3.3)

We'll require also one to one correspondence, between the two sets of variable pairs: (u,v) and (x,y), which is essential for changing variables of integration. This requirement confines the change of variables to a transformation of coordinates, studied in the previous section. This one to one transformation is called mapping, as far as points of one coordinate system are matched to the corresponding points of the other.

 $\left(x,y\right)⇔\left(u,v\right)$ (3.2.3.4)

If we express the (u,v) variables as a set of new Cartesian coordinates, in order to integrate in this plane, we have to know the magnification factor  μ  of a rectangular element of "area"  ΔuΔv, mapped to the corresponding element  ΔxΔy  from the original plane:

 $\begin{array}{l}F\left(x,y\right)\Delta x\Delta y=F\left[x\left(u,v\right),y\left(u,v\right)\right]\mu \left(u,v\right)\Delta u\Delta v\\ \text{or}\text{ }\Delta x\Delta y=\mu \left(u,v\right)\Delta u\Delta v\end{array}\right\}$ (3.2.3.5)

Since the "area" elements  ΔuΔv  and  ΔxΔy  are by the definition of a double integral - positive, (3.2.3.5) requires that the magnification factor  μ  should be non-negative.We are going to see that  μ  is the absolute value of the Jacobian  (3.1.7.7) :

 $\Delta x\Delta y=|\frac{D\left(x,y\right)}{D\left(u,v\right)}|\Delta u\Delta v=|\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}|\Delta u\Delta v$ (3.2.3.6)

and therefore the balance of the physical dimensions is kept.

## Jacobian as magnifier

In order to prove (3.2.3.6), we'll plot a  ΔuΔv  element in the original (x,y) Cartesian system. Then we'll express the corresponding "area" by use of the transformation (3.2.3.3). The result should be the the value of  $\mu \left(u,v\right)\Delta u\Delta v$  (3.2.3.5). The graphical presentation is available in Fig. Jacobian as magnifier, and it is advisable to keep the figure-window open in parallel to the present study.

Small enough elements of  Δu  and  Δv  can be approximated to straight segments, forming a quadrangle in the  x,y  plane. The evaluation of the  x,y  coordinates will be achieved by first approximation of (3.2.3.3). According to the figure, starting from the point  x,y, the additional points of the quadrangle's corners are

 $\begin{array}{ccc}i=& {x}_{i}=& {y}_{i}=\\ 1& \text{ }x+\frac{\partial x}{\partial v}\Delta v& \text{ }y+\frac{\partial y}{\partial v}\Delta v\\ 2& \text{ }x+\frac{\partial x}{\partial v}\Delta v+\frac{\partial x}{\partial u}\Delta u& \text{ }y+\frac{\partial y}{\partial v}\Delta v+\frac{\partial y}{\partial u}\Delta u\\ 3& \text{ }\text{ }x+\frac{\partial x}{\partial u}\Delta u\text{ }& \text{ }\text{ }\text{ }y+\frac{\partial y}{\partial u}\Delta u\text{ }\text{ }\end{array}\text{ }\right\}$ (3.2.3.7)

The area of the quadrangle is obtained by integrating clockwise over its sides:

 $\begin{array}{l}\frac{1}{2}\left\{\begin{array}{l}\left({x}_{1}-x\right)\left({y}_{1}+y\right)+\left({x}_{2}-{x}_{1}\right)\left({y}_{2}+{y}_{1}\right)+\\ \left({x}_{3}-{x}_{2}\right)\left({y}_{3}+{y}_{2}\right)+\left(x-{x}_{3}\right)\left(y+{y}_{3}\right)\end{array}\right\}=\\ =\frac{1}{2}\left\{\begin{array}{l}\frac{\partial x}{\partial v}\Delta v\left(2y+\frac{\partial y}{\partial v}\Delta v\right)+\frac{\partial x}{\partial u}\Delta u\left(2y+2\frac{\partial y}{\partial v}\Delta v+\frac{\partial y}{\partial u}\Delta u\right)+\\ -\frac{\partial x}{\partial v}\Delta v\left(2y+\frac{\partial y}{\partial v}\Delta v+2\frac{\partial y}{\partial u}\Delta u\right)-\frac{\partial x}{\partial u}\Delta u\left(2y+\frac{\partial y}{\partial u}\Delta u\right)\end{array}\right\}=\\ =\left(\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right)\Delta u\Delta v\end{array}\right\}$ (3.2.3.8)

Since we use the magnification in its absolute value, (3.2.3.8) yields (3.2.3.6), as required.

For example we'll recalculate the area of an ellipse, but this time by using a two dimensional-integration in a more suitable curvilinear coordinate system than the Cartesian one. Starting from the known parametric representation of an ellipse (1.2.5.37), we can form a transformation of coordinates:

 $\begin{array}{l}x=\gamma a\mathrm{cos}\beta \\ y=\gamma b\mathrm{sin}\beta \end{array}\right\}$ (3.2.3.9)

where  β  and  γ  are the new variables. The  γ  can take only non-negative values, and the interval of  β  is [0,2π). That way this transformation maps any point of one set of the coordinates, to a point in the other. The only exception is the origin of the Cartesian coordinates, which corresponds to  γ = 0 , but  β  could take any value, similarly to the polar coordinates, as studied previously. For any other value of  γ  one obtains an ellipse with the same ratio of radii (a/b), for  γ = 1 the radii are  a  and  b. Any point enclosed in this ellipse is mapped to a point enclosed in a rectangle of the new coordinate system. The "area" in the (β,γ) coordinate system corresponds to:

 $\underset{0}{\overset{1}{\int }}\text{d}\gamma \underset{0}{\overset{2\pi }{\int }}\text{d}\beta =2\pi$ (3.2.3.10)

which does not make any sense, without using the Jacobian as the magnification factor:

 $\mu \left(\beta ,\gamma \right)=|\frac{\partial x}{\partial \beta }\frac{\partial y}{\partial \gamma }-\frac{\partial x}{\partial \gamma }\frac{\partial y}{\partial \beta }|=\gamma ab|-{\mathrm{sin}}^{2}\beta -{\mathrm{cos}}^{2}\beta |=\gamma ab$ (3.2.3.11)

The transformation (3.2.3.9-11) is illustrated in Fig. Mapping transformed points. The area of the ellipse, calculated in the new coordinate system is therefore:

 $\underset{0}{\overset{1}{\int }}\text{d}\gamma \underset{0}{\overset{2\pi }{\int }}\mu \left(\beta ,\gamma \right)\text{d}\beta =ab\underset{0}{\overset{1}{\int }}\gamma \text{d}\gamma \underset{0}{\overset{2\pi }{\int }}\text{d}\beta =\frac{2\pi ab}{2}=\pi ab$ (3.2.3.12)

as expected. The volume of the ellipsoid

 $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}+\frac{{z}^{2}}{{c}^{2}}=1$ (3.2.3.13)

was calculated at (3.2.1.26) by integrating in two dimensions over this ellipse with the function

 $z=c\sqrt{1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}$ (3.2.3.14)

As an additional example we'll repeat the integration, but this time in the new coordinate system that we just used. The expression (3.2.3.14), with the aid of (3.2.3.9), becomes

 $z=c\sqrt{1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{c}^{2}}}=c\sqrt{1-{\gamma }^{2}}$ (3.2.3.15)

and by applying the magnification factor (3.2.3.11), the volume V of the ellipsoid is

 $\begin{array}{l}V=2abc\underset{0}{\overset{1}{\int }}\gamma \sqrt{1-{\gamma }^{2}}\text{d}\gamma \underset{0}{\overset{2\pi }{\int }}\text{d}\beta =4\pi abc\underset{0}{\overset{1}{\int }}{u}^{2}\text{d}u=\frac{4\pi }{3}abc\\ \text{where}\text{ }1-{\gamma }^{2}={u}^{2}\text{ }⇒\text{ }\gamma \text{d}\gamma =-u\text{d}u\end{array}\right\}$ (3.2.3.16)

These two examples show that an appropriate choice of coordinates' transformation can greatly simplify the integration.

A linear transformation is defined by a linear relation with constant factors and non-vanishing determinant:

 $\begin{array}{l}\left\{\begin{array}{l}x={a}_{11}u+{a}_{12}v\\ y={b}_{21}u+{b}_{22}v\end{array}\right\}\\ |\begin{array}{cc}{a}_{11}& {a}_{12}\\ {b}_{21}& {b}_{22}\end{array}|\ne 0\end{array}\right\}$ (3.2.3.17)

In such a case the magnification factor is a constant and does not take part in the integration itself, but is an external multiplier, simplifying that way the calculation. As an example, we'll recalculate the area of an ellipse, by simple scaling of the coordinates:

 $\begin{array}{l}\text{by}\text{\hspace{0.28em}}\text{the}\text{\hspace{0.28em}}\text{transformation}\left\{\begin{array}{l}x=au\\ y=bv\end{array}\right\}\\ \text{the}\text{\hspace{0.28em}}\text{ellipse}\text{ }\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\\ \text{becomes}\text{\hspace{0.28em}}\text{a circle}\text{ }{u}^{2}+{v}^{2}=1\\ \text{and}\text{ }\frac{D\left(x,y\right)}{D\left(u,v\right)}=\text{ab}\end{array}\right\}$ (3.2.3.18)

Since the "area" in the  u,v  coordinates is a circle with a radius of one unit, its area is  π, which multiplied by the Jacobian gives  πab.

## Polar coordinates

The known relation of the polar with the Cartesian coordinates (1.2.5.28-29) 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 the Cartesian coordinates, is given by:

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

The polar coordinates are orthogonal, since the lines corresponding to constant  r  intersect the lines of constant $\phi$ at right angles, when drawn in the Cartesian system. Therefore the quadrangle formed by the elements  Δr  and  Δ $\phi$   represents by first approximation a rectangle, allowing to obtain directly the area element (3.2.3.21). This is illustrated in Fig. Orthogonality of polar coordinates. The same result (3.2.3.21) can be obtained directly, from the approximated area  A  of a ring, whose width is  Δr, multiplied by the angular factor  (Δ $\phi$) / (2π) :

 $\begin{array}{l}A=\pi \left[{\left(r+\Delta r\right)}^{2}-{r}^{2}\right]\cong 2\pi r\Delta r\\ \frac{\Delta \phi }{2\pi }2\pi r\Delta r=r\Delta r\Delta \phi \end{array}\right\}$ (3.2.3.22)

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  f(r,φ) = 1  (3.2.3.23) expresses the area of the region:

 $\underset{{\phi }_{\mathrm{min}}}{\overset{{\phi }_{\mathrm{max}}}{\int }}\text{d}\phi \underset{0}{\overset{r\left(\phi \right)}{\int }}r\text{​}\text{d}r=\frac{1}{2}\underset{{\phi }_{\mathrm{min}}}{\overset{{\phi }_{\mathrm{max}}}{\int }}{r}^{2}\left(\phi \right)\text{d}\phi$ (3.2.3.24)

in agreement with (2.2.3.36), obtained by the use of single integrals.

We'll use a circle with radius  a  and centre at (x,y)=(0,a) as example. It is known from chapter 1 (1.2.5.39) that its representation in polar coordinates is:

$r=2a\mathrm{sin}\phi \text{ }\text{with}\text{ }\text{0}\le \phi \le \text{π}$ (3.2.3.25)

From (3.2.3.24),we obtain that its area A is

 $A=\frac{1}{2}\underset{0}{\overset{\pi }{\int }}\text{d}\phi {\left(2a\text{​}\mathrm{sin}\phi \right)}^{2}=2{a}^{2}\underset{0}{\overset{\pi }{\int }}{\mathrm{sin}}^{2}\phi \text{d}\phi =\pi {a}^{2}$ (3.2.3.26)

as expected. If the circle represents a planar body, with constant surface density of  σ, than its moment of inertia about the  x  axis, according to (3.2.3.23) is

 $\begin{array}{l}{I}_{x}=\sigma \underset{0}{\overset{\pi }{\int }}\text{d}\phi \underset{0}{\overset{2a\text{​}\mathrm{sin}\phi }{\int }}{y}^{2}\left(r,\phi \right)\text{​}r\text{d}r=\sigma \underset{0}{\overset{\pi }{\int }}{\mathrm{sin}}^{2}\phi \text{​}\text{d}\phi \underset{0}{\overset{2a\text{​}\mathrm{sin}\phi }{\int }}{r}^{3}\text{d}r=\\ =\frac{16{a}^{4}\sigma }{4}\underset{0}{\overset{\pi }{\int }}{\mathrm{sin}}^{6}\phi \text{​}\text{d}\phi =4{a}^{4}\sigma \frac{5}{6}\frac{3}{4}\frac{\pi }{2}=\frac{5}{4}M{a}^{2}\end{array}\right\}$ (3.2.3.27)

where  M  is the mass of the body. The same circle with its centre at the origin has the simple representation

 $r=a\text{ }\text{for}\text{ }0\le \phi \le 2\pi$ (3.2.3.28)

By the use of the pattern of (3.2.3.27) for this circle, we'll obtain the moment of inertia for the same body, but with the centre of mass (the centre of the circle) on the axis:

 $\begin{array}{l}{I}_{cm}=\sigma \underset{0}{\overset{2\pi }{\int }}\text{d}\phi \underset{0}{\overset{a\text{​}}{\int }}{y}^{2}\left(r,\phi \right)\text{​}r\text{d}r=\sigma \underset{0}{\overset{2\pi }{\int }}{\mathrm{sin}}^{2}\phi \text{​}\text{d}\phi \underset{0}{\overset{a}{\int }}{r}^{3}\text{d}r=\\ =\frac{{a}^{4}\sigma }{4}\underset{0}{\overset{2\pi }{\int }}{\mathrm{sin}}^{2}\phi \text{​}\text{d}\phi =\frac{\pi {a}^{4}\sigma }{4}=\frac{1}{4}M{a}^{2}\end{array}\right\}$ (3.2.3.29)

as expected from Steiner's theorem (3.2.2.34).

Another example is related to the the Gaussian function, which by definition (exercise 2) is

$f\left(x\right)=\mathrm{exp}\left(-\frac{{x}^{2}}{2}\right)$ (3.2.3.30)

At (2.2.3.12) it was shown that its improper integral of the first kind converges. There is a simple trick using polar coordinates, to calculate the integral

 $I=\underset{-\infty }{\overset{\infty }{\int }}\mathrm{exp}\left(-\frac{{x}^{2}}{2}\right)\text{​}\text{d}x$ (3.2.3.31)

which is given here as an example:

 $\begin{array}{l}{I}^{2}={\left[\underset{-\infty }{\overset{\infty }{\int }}\mathrm{exp}\left(-\frac{{x}^{2}}{2}\right)\text{d}x\right]}^{2}=\left[\underset{-\infty }{\overset{\infty }{\int }}\mathrm{exp}\left(-\frac{{x}^{2}}{2}\right)\text{d}x\right]\text{ }\left[\underset{-\infty }{\overset{\infty }{\int }}\mathrm{exp}\left(-\frac{{y}^{2}}{2}\right)\text{d}y\right]=\\ =\underset{-\infty }{\overset{\infty }{\int }}\mathrm{exp}\left(-\frac{{x}^{2}}{2}\right)\text{d}x\text{ }\underset{-\infty }{\overset{\infty }{\int }}\mathrm{exp}\left(-\frac{{y}^{2}}{2}\right)\text{d}y=\underset{-\infty }{\overset{\infty }{\int }}\text{d}x\text{ }\underset{-\infty }{\overset{\infty }{\int }}\mathrm{exp}\left(-\frac{{x}^{2}+{y}^{2}}{2}\right)\text{d}y=\\ =\underset{0}{\overset{2\pi }{\int }}\text{d}\phi \underset{0}{\overset{\infty }{\int }}r\text{ }\mathrm{exp}\left(-\frac{{r}^{2}}{2}\right)\text{d}r=2\pi \underset{0}{\overset{\infty }{\int }}\mathrm{exp}\left(-u\right)\text{​}\text{d}u=2\pi \end{array}\right\}$ (3.2.3.32)

where the transformation from Cartesian to polar coordinates was used, and the integration was done over the whole two dimensional plane. As a consequence, the normalized Gaussian function is defined as:

 $G\left(x\right)=\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-\frac{{x}^{2}}{2}\right)\text{ }\text{ }\text{with}\text{ }\text{ }\underset{-\infty }{\overset{\infty }{\int }}G\left(x\right)\text{d}x=1$ (3.2.3.33)

For a region of integration in polar coordinates, 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)

which is illustrated in Fig. Region excluding the origin.

It goes without saying that for non-convex regions, special care should be taken to accommodate the integration over the particular region, as done with Cartesian coordinates.

## Exercises

Exercise 1. Use the transformation of coordinates $\begin{array}{l}x=u\\ y={u}^{3}+u{v}^{2}+{v}^{3}\end{array}\right\}$ to answer the following questions:

1. Obtain the Jacobian $\frac{D\left(x,y\right)}{D\left(u,v\right)}$ in terms of u and v !

2. What is the corresponding "area" in the (x,y) plane, of the rectangular region in the (u,v) plane defined by
$\begin{array}{l}0\le u\le {u}_{0}\\ 0\le v\le {v}_{0}\end{array}\right\}$ where u0 and v0 are positive constants ?

3. Express the sides of the (u,v) rectangle (question 2) as curves in the (x,y) plane!

4. Make a sketch of the corresponding region in the (x,y) plane!

5. Use the results of questions 3 and 4, in order to integrate directly the "area" of this region in the (x,y) plane, without using the Jacobian!

Exercise 2. Answer the following questions by using the transformation of coordinates  $\left(x,y\right)⇔\left(\beta ,\gamma \right)$  :

$\begin{array}{l}x=\gamma a\text{ }\mathrm{cosh}\beta \\ y=\gamma b\text{ }\mathrm{sinh}\beta \end{array}$

where  a and b  are positive constants, and  x, y, a, b  have physical dimensions of length.

1. What is the domain of (x,y) subject to this transformation? (hint: $\text{tanh}\beta$)

2. Obtain the absolute value of the Jacobian $|\frac{D\left(x,y\right)}{D\left(\beta ,\gamma \right)}|$ in terms of β and γ !
3. Consider the region enclosed between the positive x branch of the hyperbola

$\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ and the straight line $x={x}_{0}>a$
Obtain the corresponding curves after transforming to $\left(\beta ,\gamma \right)$ coordinates.
4. Make a sketch of the regions in question 3 !
5. Consider the region (of question 3) as a planar body with density (mass per area)

$\sigma ={\sigma }_{0}\frac{\frac{x}{{x}_{0}}}{\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}$ where ${\sigma }_{0}$ is a positive constant.
Use the $\left(\beta ,\gamma \right)$ variables, in order to calculate the mass M of the body!
Check the physical dimensions!

Exercise 3. Answer the following questions by using the transformation of coordinates  $\left(x,y\right)⇔\left(\beta ,\gamma \right)$

$\begin{array}{l}x=\gamma a\text{ }\mathrm{cos}\beta \\ y=\gamma b\text{ }\mathrm{sin}\beta \end{array}$
where  a and b  are positive constants, and  x, y, a, b  have physical dimensions of length.

1. The density (mass/area) of the uniform planar solid ellipse

$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$
is  σ . What is the moment of inertia  Iy  about the y axis?
2. As a continuation of question 1, two circular holes with radius  ρ  each contained fully by the ellipse are perforated. The centres of the holes are located at  (x,y)=(±d,0), where  d>ρ . What is the new moment of inertia? (Hint: Steiner's theorem)

3. Without relation to the questions 1 and 2, what is the volume enclosed between the partial surface of the elliptic paraboloid

$z=h\left(1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}\right)\ge 0$
and the (x,y) plane, where h is a positive constant? Compare this with the results for the same volume done in a previous exercise by a different method.
4. Check the final results of questions 1,2,3 for consistency with physical dimensions!

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 as  r = r(φ) !
2. Calculate the area  A  of the planar body!
3. What is the x coordinate of its centre of mass?

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