I have the following transforms:
$\begin{align} x &= u^2 - v^2 \\ y &= 2uv \end{align}$
and am tasked with finding the area between the following curves:
$\begin{align} x &= 4 - \frac{y^2}{16} \\ x &= \frac{y^2}{4} - 1 \end{align}$
So, I substituted the transforms into the equations for the curves:
$\begin{align} u^2 - v^2 &= 4 - \frac{4u^2v^2}{16} \\ u^2 - v^2 &= \frac{4u^2v^2}{14} - 1 \end{align}$
Solving the first for $u$ and the second for $v$, I get:
$\begin{align} u &= \pm2 \\ v &= \pm1 \end{align}$
And figure the area between these two curves must be:
$\begin{align} 4 \int_{-1}^1 \int_{-2}^2 (u^2 + v^2) \,du \,dv \end{align}$
Solving this, I end up at $\frac{160}{3}$; my books says I should end up at $\frac{50}{3}$
Am I misunderstanding how to use transforms and the Jacobian to find the area between two curves, or does it look like I've made a mistake somewhere in my algebra?
\begin{align*} x &= \left( \frac{y}{2v} \right)^{2}-v^{2} \\ &= u^{2}-\left( \frac{y}{2u} \right)^{2} \end{align*}
Note that $(u,v) \in [0,2] \times [0,1]$ (blue and red grids) has already covered half of the area (green) below:
Avoid double count, either take $(u,v) \in [0,2] \times [-1,1]$ or $[-2,2] \times [0,1]$
\begin{align*} dA &= 4(u^2+v^2) \, du \, dv \\ A &= \int_{-1}^{1} \int_{0}^{2} 4(u^2+v^2) \, du \, dv \\ &= \int_{-1}^{1} \left[ \frac{4u^{3}}{3}+4uv^{2} \right]_{u=0}^{2} dv \\ &= \int_{-1}^{1} \left( \frac{32}{3}+8v^2 \right) dv \\ &= \left[ \frac{32v}{3}+\frac{8v^{3}}{3} \right]_{v=-1}^{1} \\ &= \frac{80}{3} \end{align*}