Mathematical Introduction for Physics and Engineering

by Samuel Dagan (Copyright © 2007-2020)

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

- Obtain the Jacobian
$\frac{D\left(x,y\right)}{D\left(u,v\right)}$
in terms of
*u*and*v*! - What is the corresponding "area" in the (
*x*,*y*) plane, of the rectangular region in the (*u*,*v*) plane defined by

$$\begin{array}{l}0\le u\le {u}_{0}\\ 0\le v\le {v}_{0}\end{array}\}$$ where*u*_{0}and*v*_{0}are positive constants ? - Express the sides of the (
*u*,*v*) rectangle (question 2) as curves in the (*x*,*y*) plane! - Make a sketch of the corresponding region in the (
*x*,*y*) plane! - 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!

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

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

$$\begin{array}{l}F\left(x,y\right)\Delta x\Delta y=F\left[x\left(u,v\right),y\left(u,v\right)\right]\mu \left(u,v\right)\Delta u\Delta v\\ \text{or}\text{\hspace{1em}}\Delta x\Delta y=\mu \left(u,v\right)\Delta u\Delta v\end{array}\}$$ | (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

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

and therefore the balance of the physical dimensions is kept.

Solution of question 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

$$\frac{\partial x}{\partial u}=1\text{\hspace{1em}}\text{and}\text{\hspace{1em}}\frac{\partial x}{\partial v}=0$$

Solution of question 2.

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

$$A=\int \int \text{d}x\text{\hspace{0.05em}}\text{d}y=\int \int \left(2uv+3{v}^{2}\right)\text{d}u\text{\hspace{0.05em}}\text{d}v$$ - The limits of integration yield the "area"

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

Solution of question 3.

- One of the rectangle's side is the line

$$u=0\text{\hspace{1em}}\left(\text{with}\text{\hspace{1em}}0\le v\le {v}_{0}\right)$$

which after substitution in the transformation

$$\begin{array}{l}x=u\\ y={u}^{3}+u{v}^{2}+{v}^{3}\end{array}\}$$ yields $$x=0\text{\hspace{1em}}\left(\text{with}\text{\hspace{1em}}0\le y\le {v}_{0}^{3}\right)$$ - Another side is

$$v=0\text{\hspace{1em}}\left(\text{with}\text{\hspace{1em}}0\le u\le {u}_{0}\right)$$

yielding $$y={x}^{3}\text{\hspace{1em}}\left(\text{with}\text{\hspace{1em}}0\le x\le {u}_{0}\right)$$ - The third side is

$$u={u}_{0}\text{\hspace{1em}}\left(\text{with}\text{\hspace{1em}}0\le v\le {v}_{0}\right)$$

which yields $$x={u}_{0}\text{\hspace{1em}}\left(\text{with}\text{\hspace{1em}}{u}_{0}^{3}\le y\le {u}_{0}^{3}+{u}_{0}{v}_{0}^{2}+{v}_{0}^{3}\right)$$ - Finally the side

$$v={v}_{0}\text{\hspace{1em}}\left(\text{with}\text{\hspace{1em}}0\le u\le {u}_{0}\right)$$

yields $$y={x}^{3}+x{v}_{0}^{2}+{v}_{0}^{3}\text{\hspace{1em}}\left(\text{with}\text{\hspace{1em}}0\le x\le {u}_{0}\right)$$

Solution of question 4.

- See the
**Fig. Sketch for exercise 1**.

Solution of question 5.

- 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.

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.