Area in R2 bounded by 4 curves What's the area between:
$$y = 5x$$
$$y = 15x$$
$$xy = 8$$
$$xy = 4$$
?
I calculated all the points of intersection and got a double integral that looks something like this: $$\int_{\frac{2}{\sqrt15}}^{\frac{2\sqrt 2}{\sqrt 5}} \int_{5x}^{15x} 1 \, \mathrm dx \, \mathrm dy$$
This doens't seem right but I'm not sure why. Any help would welcome!
 A: You can compute the area by a change of variables 
$$(u,v) = \left(xy,\frac{y}{x}\right)$$
To compute the integral in new coordinates, you need the value of the
Jacobian determinant:
$$\frac{\partial(x,y)}{\partial(u,v)}\quad\stackrel{def}{=}\quad
\left|\begin{matrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{matrix}\right|
$$
Instead of computing this explictly, one can obtain it by manipulatiing the "area element" in an algebraic manner. We have
$$dx \wedge dy = dx \wedge d(xv) = x dx \wedge dv = \frac12 d(x^2) \wedge dv
= \frac12 d\frac{u}{v} \wedge dv = \frac{du \wedge dv}{2v}$$
This implies up to a sign, the Jacobian determinant is
$$\frac{\partial(x,y)}{\partial(u,v)} = \frac{1}{2v}$$
In the new coordinates $(u,v)$, the area becomes an integral over a rectangle
$[4,8] \times [5,15]$, i.e.
$$\begin{align}
\verb/Area/ 
&=  \int_{[4,8]\times[5,15]} \left|\frac{\partial(x,y)}{\partial(u,v)}\right| du dv
=  \frac12 \int_4^8 \left[ \int_5^{15} \frac{dv}{v}\right] du\\ 
&= \frac12 (8-4)\big[ \log 15 - \log 5 \big] = 2\log 3
\end{align}$$
A: I try to draw the above curves using matlab, and  the enclosed area can be found as follow: $$A=\int_{\sqrt\frac{4}{15}}^{\sqrt\frac{8}{15}} \int_{\frac{4}{x}}^{15x}1 dydx + \int_{\sqrt\frac{8}{15}}^{\sqrt\frac{4}{5}} \int_{\frac{4}{x}}^{\frac{8}{x}}1 dydx  + \int_{\sqrt\frac{4}{5}}^{\sqrt\frac{8}{5}} \int_{5x}^{\frac{8}{X}}1 dydx  $$
