Find area of region bounded by $xy=4, xy=8, xy^3=5, xy^3=15$.
Can someone help me check if there's anything wrong with my working or conceptual understanding of it? The answer in my textbook is $2\ln3$ but my answer is $\frac{8}{3}\ln3$.
Let $xy=u$ and $xy^3=v$
Calculating change of integrals:
$\frac{\partial x}{\partial u}=\frac{1}{y}$ $\frac{\partial y}{\partial u}=\frac{1}{x}$ $\frac{\partial x}{\partial v}=\frac{1}{y^3}$ $\frac{\partial y}{\partial v}=\frac{1}{3y^2x}$
So, $dydx= (\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v}) dudv$
$dydx= \frac{1}{3y^3x}-\frac{1}{y^3x}dudv$
$dydx= -\frac{2}{3y^3x}dudv$
$dydx= -\frac{2}{3v}dudv$
$\int^{15}_5 \int^8 _4 -\frac{2}{3v} dudv = -\frac{8}{3}\ln3$
Area = $\frac{8}{3}\ln3$
When you are taking derivative with respect to $u$ (or $v$) you should apply it to both $x$ and $y$, as they are dependent on it. To avoid this problem, express $x$ and $y$ in terms of $u,v$ (in particular $x^2= \frac{u^3}{v}$ and $y^2 = \frac vu$) or an even better solution is to calculate the Jacobian of the inverse trasformation and take the reciprocal value. We have:
$$\frac{\partial u}{\partial x} = y; \frac{\partial u}{\partial y} = x; \frac{\partial v}{\partial x} = y^3; \frac{\partial v}{\partial y} = 3xy^2$$ Then:
$$\left|\begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} \\ \end{array}\right| = \left|\begin{array}{cc} y & y^3 \\ x & 3xy^2 \\ \end{array}\right| = 3xy^3-xy^3 = 2xy^3 = 2v$$
Thus your Jacobian is $\frac{1}{2v}$. Now to find the area:
$$\int_5^{15}\int_4^8 \frac{1}{2v} dudv = 2\int_{5}^{15} \frac 1v dv = 2 \ln v \; \bigg|_5^{15} = 2 \ln\left( \frac{15}{5}\right) = 2 \ln 3$$