]> Exercise 1

# Two Dimensional Transformation, Exercise 1

## Question

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!

## Reminder

... The change of two variables ... 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 ....

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 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.

## Part 1

Solution of question 1.

1. The Jacobian (3.2.3.6) is

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

## Parts 2-3

Solution of question 2.

1. Since the Jacobian is non-negative in the region, the required "area" A should be expressed by

$A=\int \int \text{d}x\text{ }\text{d}y=\int \int \left(2uv+3{v}^{2}\right)\text{d}u\text{ }\text{d}v$

2. The limits of integration yield the "area"

$\begin{array}{l}A=\underset{0}{\overset{{u}_{0}}{\int }}\text{d}u\underset{0}{\overset{{v}_{0}}{\int }}\text{d}v\left(2uv+3{v}^{2}\right)=\underset{0}{\overset{{u}_{0}}{\int }}\text{d}u\left(u{v}_{0}^{2}+{v}_{0}^{3}\right)=\\ =\frac{1}{2}{u}_{0}^{2}{v}_{0}^{2}+{u}_{0}{v}_{0}^{3}={u}_{0}{v}_{0}^{2}\left(\frac{{u}_{0}}{2}+{v}_{0}\right)\end{array}$

## Parts 4-7

Solution of question 3.

1. One of the rectangle's side is the line

$u=0\text{ }\left(\text{with}\text{ }0\le v\le {v}_{0}\right)$
which after substitution in the transformation

$\begin{array}{l}x=u\\ y={u}^{3}+u{v}^{2}+{v}^{3}\end{array}\right\}$ yields $x=0\text{ }\left(\text{with}\text{ }0\le y\le {v}_{0}^{3}\right)$

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

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

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

## Part 8

Solution of question 4.

1. See the Fig. Sketch for exercise 1.

## Part 9

Solution of question 5.

1. According the results of question 3, together with the sketch, the "area" can be integrated directly:

$\begin{array}{l}A=\underset{0}{\overset{{u}_{0}}{\int }}\text{d}x\underset{{y}_{\mathrm{min}}\left(x\right)}{\overset{{y}_{\mathrm{max}}\left(x\right)}{\int }}\text{d}y=\underset{0}{\overset{{u}_{0}}{\int }}\text{d}x\left[{y}_{\mathrm{max}}\left(x\right)-{y}_{\mathrm{min}}\left(x\right)\right]=\\ =\underset{0}{\overset{{u}_{0}}{\int }}\text{d}x\left[\left({x}^{3}+x{v}_{0}^{2}+{v}_{0}^{3}\right)-{x}^{3}\right]=\\ =\underset{0}{\overset{{u}_{0}}{\int }}\text{d}x\left(x{v}_{0}^{2}+{v}_{0}^{3}\right)={u}_{0}{v}_{0}^{2}\left(\frac{{u}_{0}}{2}+{v}_{0}\right)\end{array}$

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.