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 1

Question

Reminder

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

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 ....

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:

Since the "area" elements ΔuΔv and ΔxΔy are by 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) :

and therefore the balance of the physical dimensions is kept.

Part 1

Solution of question 1.

1. The Jacobian (3.2.3.6) is
   
   $$\frac{D\left(x,y\right)}{D\left(u,v\right)}=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}=\frac{\partial y}{\partial v}=2uv+3v^2$$
   since
   $$\frac{\partial x}{\partial u}=1\text{and}\frac{\partial x}{\partial v}=0$$

Parts 2-3

Solution of question 2.

2. Since the Jacobian is non-negative in the region, the required "area" A should be expressed by
   
   $$A=\int \int \text{d}x\text{d}y=\int \int \left(2uv+3v^2\right)\text{d}u\text{d}v$$


3. The limits of integration yield the "area"
   
   $$\begin{array}{l}A=\underset{0}{\overset{u_0}{\int }}\text{d}u\underset{0}{\overset{v_0}{\int }}\text{d}v\left(2uv+3v^2\right)=\underset{0}{\overset{u_0}{\int }}\text{d}u\left(u{v}_0^2+v_0^3\right)=\\ =\frac{1}{2}{u}_0^2{v}_0^2+u_0{v}_0^3=u_0{v}_0^2\left(\frac{u_0}{2}+v_0\right)\end{array}$$

Parts 4-7

Solution of question 3.

4. One of the rectangle's side is the line
   
   $$u=0\left(\text{with}0\le v\le v_0\right)$$
   which after substitution in the transformation
   
   $$\begin{array}{l}x=u\\ y=u^3+u{v}^2+v^3\end{array}\}$$ yields $$x=0\left(\text{with}0\le y\le v_0^3\right)$$


5. Another side is
   $$v=0\left(\text{with}0\le u\le u_0\right)$$
   yielding $$y=x^3\left(\text{with}0\le x\le u_0\right)$$


6. The third side is
   $$u=u_0\left(\text{with}0\le v\le v_0\right)$$
   which yields $$x=u_0\left(\text{with}{u}_0^3\le y\le u_0^3+u_0{v}_0^2+v_0^3\right)$$


7. Finally the side
   $$v=v_0\left(\text{with}0\le u\le u_0\right)$$
   yields $$y=x^3+x{v}_0^2+v_0^3\left(\text{with}0\le x\le u_0\right)$$

Part 8

Solution of question 4.

8. See the Fig. Sketch for exercise 1.

Part 9

Solution of question 5.

9. According the results of question 3, together with the sketch, the "area" can be integrated directly:
   
   $$\begin{array}{l}A=\underset{0}{\overset{u_0}{\int }}\text{d}x\underset{y_{\min }\left(x\right)}{\overset{y_{\max }\left(x\right)}{\int }}\text{d}y=\underset{0}{\overset{u_0}{\int }}\text{d}x\left[y_{\max }\left(x\right)-y_{\min }\left(x\right)\right]=\\ =\underset{0}{\overset{u_0}{\int }}\text{d}x\left[\left(x^3+x{v}_0^2+v_0^3\right)-x^3\right]=\\ =\underset{0}{\overset{u_0}{\int }}\text{d}x\left(x{v}_0^2+v_0^3\right)=u_0{v}_0^2\left(\frac{u_0}{2}+v_0\right)\end{array}$$
   
   The result is identical to that of part 3, as it should be.

Score

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

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