Double Integrals, Jacobians, Thermodynamics

[_N3WTON_]

Homework Statement

An important problem in thermodynamics is to find the work done by an Ideal Carnot engine. A cycle consists of alternating expansion and compression of gas in a piston. The work done by the engine is equal to the area of the region R enclosed by two isothermal curves $xy = a$, $xy = b$ and two adiabatic curves $xy^{1.4} = c$, $xy^{1.4} = d$, where $0 < a < b$ and $0 < c < d$. Compute the work done by determining the area of R.

Homework Equations

$\int \int f(x,y),dA = \int \int f(x(u,v),y(u,v))\frac{\partial (x,y)}{\partial (u,v)} du dv$
$\frac{\partial (x,y)}{\partial (u,v)} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}$

The Attempt at a Solution

First, transforming to u and v:
$u = xy$
$v = xy^{1.4}$
Now, finding the inverse transformation by solving the above equations:
$x = \frac{u}{y}$
$v = \frac{u}{y}y^{1.4}$
$v = uy^{\frac{2}{5}}$
$y^{\frac{2}{5}} = \frac{v}{u}$
$y = (\frac{v}{u})^{\frac{5}{2}}$
I plugged this value into my equation for x, I won't show the work because it is hard for me to write in latex :D. This is my result:
$x = \frac{u^{\frac{7}{2}}}{v^{\frac{5}{2}}}$
$x(u,v) = \frac{u^{\frac{7}{2}}}{v^{\frac{5}{2}}}$
$y(u,v) = \frac{v^{\frac{5}{2}}}{u^{\frac{5}{2}}}$
Next, I took the partial derivatives to find the Jacobian:
$\frac{\partial x}{\partial u} = \frac{7u^{\frac{5}{2}}}{2v^{\frac{5}{2}}}$
$\frac{\partial y}{\partial u} = -\frac{5v^{\frac{5}{2}}}{2u^{\frac{7}{2}}}$
$\frac{\partial x}{\partial v} = -\frac{5u^{\frac{7}{2}}}{2v^{\frac{7}{2}}}$
$\frac{\partial y}{\partial v} = \frac{5v^{\frac{3}{2}}}{2u^{\frac{5}{2}}}$
Then, using the formula to find the Jacobian (which I also won't include here because it would be hard to write in Latex) I found:
$\frac{35}{4v} - \frac{25}{4v}$
$= \frac{5}{2v}$
I know that since the problem is asking for area, the value of the function being integrated is equal to 1. So I obtained the following double integral:
$\int_{c}^{d} \int_{a}^{b} \frac{5}{2v} du \hspace{1 mm} dv$
$\int_{c}^{d} \frac{5b-5a}{2v}\,dv$
$\frac{5b-5a}{2} \int_{c}^{d} \frac{1}{v}\,dv$
So my final answer is:
$\frac{5b-5a}{2}(ln(d)-ln(c))$
I was hoping someone here would be kind enough to check my work. I asked a few people in the chat if this was bad form (asking to have your work checked) and I was told that it was not, However, if it is frowned upon please let me know. Thanks :)

 

[mighty2000]

I can confirm that it is completely acceptable to have your work checked by others. In fact, it is encouraged in the scientific community to have peer review and collaboration in order to improve and validate scientific findings.

In terms of your solution, it looks like you have followed the correct steps and your final answer seems to be correct. However, I would suggest double-checking your calculations and make sure you have accounted for all the terms in the Jacobian. Also, it might be helpful to include units in your final answer, as it represents the area of the region R.

Overall, great job on your solution and keep up the good work!