Transformation (?) of continuous random variables Background
Suppose that we are using a simplified spherical model of the Earth's surface with latitude $u \in (-\frac {\pi} 2, \frac {\pi} 2)$ and longitude $v \in (-\pi, \pi)$; then (if the radius is taken to be $1$), the surface area element is given by $\mathrm{d}A = \cos u\ \mathrm{d}u\ \mathrm{d}v$. Restricting attention to the hemisphere, $H$, where $u, v \in (-\frac {\pi} 2, \frac {\pi} 2)$, a simple map projection from $H$ can be obtained by just taking the $x$ and $y$ coordinates via $x = \cos u \sin v$ and $y = \sin u$, which is a smooth one-to-one transformation on $H$. Now, picking a point with coordinates $(U, V)$ on $H$ uniformly according to surface area, the joint density of $U$ and $V$ is $$f_{U, V}(u, v) = \frac 1 {2\pi} \cos u, \quad \lvert u \rvert, \lvert v \rvert < \frac {\pi} 2.$$
Question
Find the joint density of $X$ and $Y$, where $(X, Y)$ is the image of the random point $(U, V)$ under the map projection defined above.
My working
Some preliminary results:
$$\begin{aligned}
J & = \begin{vmatrix} \frac {dx} {du} & \frac {dx} {dv} \\ \frac {dy} {du} & \frac {dy} {dv} \end{vmatrix}
\\[1 mm] & = \begin{vmatrix} -\sin u \sin v & \cos u \cos v \\ \cos u & 0 \end{vmatrix}
\end{aligned}$$
Thus,
$$\begin{aligned}
\\[1 mm] f_{X, Y}(x, y) & = \frac 1 {2\pi} \cos u\ \lvert J \rvert
\\[1 mm] & = \frac 1 {2\pi} \cos^3 u \cos v
\\[1 mm] & = \frac 1 {2\pi} x(1 - y^2) \cot v
\\[1 mm] & = \frac 1 {2\pi} (1 - y^2) \sqrt {1 - y^2 - x^2}
\end{aligned}$$

I am stuck here. Firstly, may I know if my expression for $f_{X, Y}(x, y)$ is correct? Secondly, if it is correct, how can I obtain the support for $f_{X, Y}(x, y)$?
 A: We have$$f_{X,Y}(x,y)=f_{U,V}(u(x,y),v(x,y))\color{red}{/|}J\color{red}|$$where $J=\frac{\partial(x,y)}{\partial(u,v)}$. I have typed in red the parts where you have faltered. You have to divide by the absolute value of the Jacobian if you calculate it as $\frac{\partial(x,y)}{\partial(u,v)}$.
Thus $$f_{X,Y}(x,y)=\frac{\cos(u(x,y))}{2\pi\cos^2(u(x,y))|\cos(v(x,y))|}$$and since $v\in(-\pi/2,\pi/2),\cos v>0$, we get$$f_{X,Y}(x,y)=\frac1{2\pi\cos(u(x,y))\cos(v(x,y))}=\frac1{2\pi[\cos u\cos v](x,y)}$$and you obtained $\cos v= \frac {\sqrt {\cos^2 u - x^2}} {\cos u}$ or $\cos v \cos u= \sqrt{ \cos^2 u - x^2} =\sqrt{1-y^2-x^2}$, we get$$f_{X,Y}(x,y)=\frac1{2\pi\sqrt{1-x^2-y^2}}$$over the circle $x^2+y^2<1$ and zero otherwise. The support was obtained using the constraints on $u,v$ in $f_{U,V}$:
$$\begin{align*}
|u|<\pi/2&\implies y=\sin u\in(-1,1)\\
|v|<\pi/2&\implies \frac x{\cos u}=\frac x{\sqrt{1-y^2}}=\sin v\in(-1,1)\\
&\implies\left|\frac x{\sqrt{1-y^2}}\right|<1\\
&\implies x^2<1-y^2
\end{align*}$$
