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 2

Question

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

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 $\left|\frac{D\left(x,y\right)}{D\left(\beta ,\gamma \right)}\right|$ 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!

Reminder

... The change of two variables ... is expressed by:

We'll require also one to one correspondance, 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 ....

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:

... the magnification factor μ ..... is the absolute value of the Jacobian :

and therefore the balance of the physical dimensions is kept.

Part 1

Solution of question 1.

1. The transformation yields
   $$\frac{y}{x}=\frac{b}{a}\tanh \beta$$
   but since $$-1<\tanh \beta <1$$ we obtain $$-\frac{b}{a}<\frac{y}{x}<\frac{b}{a}$$

Part 2

Solution of question 2.

2.   $$\begin{array}{l}\left|\frac{D\left(x,y\right)}{D\left(\beta ,\gamma \right)}\right|=\left|\frac{\partial x}{\partial \beta }\frac{\partial y}{\partial \gamma }-\frac{\partial x}{\partial \gamma }\frac{\partial y}{\partial \beta }\right|=\\ =\left|ab\gamma \left({\sinh }^2\beta -{\cosh }^2\beta \right)\right|=ab\left|\gamma \right|\end{array}$$

Parts 3-4

Solution of question 3.

3. The substitution of $$\begin{array}{l}x=\gamma a\cosh \beta \\ y=\gamma b\sinh \beta \end{array}$$ into $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ yields $${\gamma }^2\left({\cosh }^2\beta -{\sinh }^2\beta \right)={\gamma }^2=1$$
   For positive x, one have to use the positive value of γ , yielding finally
   
   $$\gamma =1$$


4. the substitution of $$x=x_0$$ into $$x=\gamma a\cosh \beta$$ yields $$\gamma =\frac{x_0}{a}\frac{1}{\cosh \beta }$$ with $$\frac{x_0}{a}>1$$ as required.

Part 5

Solution of question 4.

5. See the Fig. Sketch for exercise 2.

Parts 6-9

Solution of question 5.

6. The density expressed by the $\left(\beta ,\gamma \right)$ variables is
   
   $$\sigma \text{=}{\sigma }_{\text{0}}\frac{\frac{x}{x_0}}{\frac{x^2}{a^2}-\frac{y^2}{b^2}}=\frac{{\sigma }_{\text{0}}\gamma a\cosh \beta }{x_0{\gamma }^2}=\frac{{\sigma }_{\text{0}}a}{x_0}\frac{\cosh \beta }{\gamma }$$
   and the integral of the mass is
   
   $$\begin{array}{l}M=\underset{{\beta }_{\min }}{\overset{{\beta }_{\max }}{\int }}\text{d}\beta \underset{{\gamma }_{\min }\left(\beta \right)}{\overset{{\gamma }_{\max }\left(\beta \right)}{\int }}\left|\frac{D\left(x,y\right)}{D\left(\beta ,\gamma \right)}\right|\sigma \left(\beta ,\gamma \right)\text{d}\gamma \text{=}\\ \text{=}ab\frac{{\sigma }_{\text{0}}a}{x_0}\underset{{\beta }_{\min }}{\overset{{\beta }_{\max }}{\int }}\text{d}\beta \underset{{\gamma }_{\min }\left(\beta \right)}{\overset{{\gamma }_{\max }\left(\beta \right)}{\int }}\gamma \frac{\cosh \beta }{\gamma }\text{d}\gamma \text{=}\\ \text{=}\frac{{\sigma }_{\text{0}}{a}^{2}b}{x_0}\underset{{\beta }_{\min }}{\overset{{\beta }_{\max }}{\int }}\cosh \beta \text{d}\beta \underset{{\gamma }_{\min }\left(\beta \right)}{\overset{{\gamma }_{\max }\left(\beta \right)}{\int }}\text{d}\gamma \end{array}$$


7. The limits of the integral over γ are according to question 3
   
   $$\{\begin{array}{l}{\gamma }_{\min }=1\\ {\gamma }_{\max }=\frac{x_0}{a}\frac{1}{\cosh \beta }\text{with}\frac{x_0}{a}>1\end{array}$$
   The limits for the integration over β are obtained from the intersection of  ${\gamma }_{\max }$  with  ${\gamma }_{\min }$
   
   $$\begin{array}{l}\cosh {\beta }_0=\frac{x_0}{a}\\ \{\begin{array}{l}{\beta }_{\min }=-{\beta }_0=-\text{acosh}\left(\frac{x_0}{a}\right)\\ {\beta }_{\max }={\beta }_0=\text{acosh}\left(\frac{x_0}{a}\right)\end{array}\end{array}$$


8. The solution of the integral of part 6 is therefore
   
   $$\begin{array}{l}M=\frac{{\sigma }_{\text{0}}{a}^{2}b}{x_0}\underset{-{\beta }_0}{\overset{{\beta }_0}{\int }}\cosh \beta \left(\frac{x_0}{a}\frac{1}{\cosh \beta }-1\right)\text{d}\beta =\\ =\frac{{\sigma }_{\text{0}}{a}^{2}b}{x_0}\underset{-{\beta }_0}{\overset{{\beta }_0}{\int }}\text{d}\beta \left(\frac{x_0}{a}-\cosh \beta \right)=\\ =\frac{{\sigma }_{\text{0}}{a}^{2}b}{x_0}\left(\frac{2{\beta }_0{x}_0}{a}-2\sinh {\beta }_0\right)=\\ =\frac{2{\sigma }_{\text{0}}{a}^{2}b}{x_0}\left(\frac{{\beta }_0{x}_0}{a}-\sqrt{{\cosh }^2{\beta }_0-1}\right)=\\ =\frac{2{\sigma }_{\text{0}}{a}^{2}b}{x_0}\left(\frac{{\beta }_0{x}_0}{a}-\sqrt{\frac{x_0^2}{a^2}-1}\right)=\\ =2{\sigma }_{\text{0}}ab\left(\text{acosh}\left(\frac{x_0}{a}\right)-\sqrt{1-\frac{a^2}{x_0^2}}\right)\end{array}$$


9. The expression inside the brackets is dimensionless. The dimensions of σ₀ are mass/area and of the product ab - area. Therefore the physical dimensions are balanced.

Score

By parts.
Parts 1,2,3,4,6,7,8,9 (all except 5) are worth 1 point each.
Part 5 is worth 2 points.

By questions.
Questions 1 and 2 are worth 1 point each.
Questions 3 and 4 are worth 2 points each.
Question 5 is worth 4 points.