# Double Integrals, Jacobians, Thermodynamics

1. Nov 9, 2014

### _N3WTON_

1. The problem statement, all variables and given/known data
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.

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

3. 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:
$\int_{c}^{d} \int_{a}^{b} \frac{5}{2v} du \hspace{1 mm} dv$
$\int_{c}^{d} \frac{5b-5a}{2v}\,dv$
$\frac{5b-5a}{2} \int_{c}^{d} \frac{1}{v}\,dv$
$\frac{5b-5a}{2}(ln(d)-ln(c))$
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 :)

2. Nov 14, 2014