I have a projective plane, that means all the points $(x, y, z)$ gets projected to $(X=x/z, Y=y/z, 1)$.
I would like to know what is the area of the unit sphere $x^2+y^2+z^2=1$ that gets projected to a rectangle. Let's say $1<X<2$ and $2<Y<3$ in my projective plane.
When using spherical coordinates ($x=\sin\theta \cos\varphi$, $y=\sin\theta \sin\varphi$, $z=\cos\theta$) I have $$A= \int_\varphi\int_\theta \sin \theta \, \mathrm{d}\theta \mathrm{d}\varphi$$
I would have to introduce kind of substitution that would map $(\theta, \varphi)$ to $(X,Y)$ in the way that $X=x/z=\sin\theta \cos \varphi / \cos\theta$, $Y=y/z=\sin\theta \sin \varphi / \cos\theta$ that leads, if I computed Jacobian correctly, $$A=\int_1^2 \int_2^3 \cos^3 (\arctan \sqrt{X^2+Y^2})\, \mathrm{d}X \mathrm{d}Y$$
And further using formula $\cos (\arctan x) = 1/\sqrt{1+x^2}$ I have that
$$A=\int_1^2 \int_2^3 \frac{1}{ (1+X^2+Y^2)^\frac{3}{2}}\, \mathrm{d}X \mathrm{d}Y$$
Wolfram alpha gave me the result: $$\int \int \frac{1}{ (1+X^2+Y^2)^\frac{3}{2}}\, \mathrm{d}X \mathrm{d}Y = c_1 Y + c_2 - 0.5 i \ln\frac{X \sqrt{X^2 + Y^2 + 1} + X^2 - i Y + 1}{X (-1 + i Y)} + 0.5 i \ln\frac{X \sqrt{X^2 + Y^2 + 1} + X^2 + i Y + 1}{X (-1 - i Y)}$$
Apparently the result is correct as the integration over $\mathbb{R}^2$ gives $2\pi$. Unfortunately I have to compute a bit myself in order to get definite integral and closed formula for any rectangle in $\mathbb{R}^2$.
Do you see some trick how to do this easily?
I think this is more easily done with spherical geometry than with integration.
You’re projecting antipodes to the same point, so the rectangle in the plane corresponds to two antipodal spherical quadrangles. We can calculate the area of one of them and multiply by $2$.
The sides of the rectangle correspond to segments of great circles on the unit sphere. The normal vectors of the great circles corresponding to $X=1$, $Y=2$, $X=2$ and $Y=3$ are $(1,0,-1)$, $(0,1,-2)$, $(-1,0,2)$ and $(0,-1,3)$, respectively (where the signs are chosen for an oriented traversal of the perimeter of the rectangle). Denote them by $n_i$ (with $n_0=n_4$). Then the internal angles of a rectangle formed by these great cirles are $\pi-\arccos\frac{n_i\cdot n_{i-1}}{|n_i||n_{i-1}|}$, so the area of the spherical quadrangle they form is (see Wikipedia)
\begin{eqnarray} && 2\pi-\sum_{i=1}^4\arccos\frac{n_i\cdot n_{i-1}}{|n_i||n_{i-1}|} \\&=& 2\pi-\left( \arccos\frac2{\sqrt{2\cdot5}}+\arccos\frac{-4}{\sqrt{5\cdot5}}+\arccos\frac6{\sqrt{5\cdot10}}+\arccos\frac{-3}{\sqrt{10\cdot2}}\right) \\ &\approx& 0.0353\;. \end{eqnarray}
So the total area projected onto that rectangle is twice this, about $0.0706$.