Find the solid angle subtended at the origin
Find the solid angle subtended at the origin by the rectangle $0<y<b,\,0<z<c,\,x=a>0$. Equivalently, find the area of the part of the unit sphere $x^2+y^2+z^2=1$ onto which the rectangle projects from the origin.
Explanation of Method
Projecting the rectangle onto the unit sphere from the origin, the solid-angle element at a point $(a,y,z)$ works out to $\dfrac a{(a^2+y^2+z^2)^{3/2}}\,dy\,dz$ — the standard formula for the differential solid angle subtended by a small patch of a plane at perpendicular distance $a$. Two successive elementary integrations (first in $y$, then in $z$) reduce this to a single closed-form arctangent.
- Set up the solid-angle integral.
- Integrate over $y$.
- Integrate over $z$ to obtain the final closed form.
Step 1 · Setup
$$ \Omega = \int_0^c\int_0^b\frac a{(a^2+y^2+z^2)^{3/2}}\,dy\,dz. $$
Step 2 · Integrate over $y$
With $A^2=a^2+z^2$ and $\displaystyle\int\frac{dy}{(A^2+y^2)^{3/2}}=\frac y{A^2\sqrt{A^2+y^2}}$:
$$ \int_0^b\frac{a\,dy}{(A^2+y^2)^{3/2}} = \frac{ab}{(a^2+z^2)\sqrt{a^2+b^2+z^2}}. $$
Step 3 · Integrate over $z$
With $B^2=a^2+b^2$ and the standard formula $\displaystyle\int\frac{dz}{(a^2+z^2)\sqrt{B^2+z^2}}=\frac1{a\sqrt{B^2-a^2}}\arctan\frac{z\sqrt{B^2-a^2}}{a\sqrt{B^2+z^2}}$ (here $\sqrt{B^2-a^2}=b$):
$$ \int_0^c\frac{dz}{(a^2+z^2)\sqrt{a^2+b^2+z^2}} = \frac1{ab}\arctan\frac{bc}{a\sqrt{a^2+b^2+c^2}}. $$
Combining (the $ab$ factors cancel):
$$ \Omega = \arctan\frac{bc}{a\sqrt{a^2+b^2+c^2}}. $$
Additional Notes
- This derivation matches the source exactly.
- As a sanity check, as $a\to0^+$ (with $b,c$ fixed), $\Omega\to\arctan(+\infty)=\frac\pi2$: the rectangle flattens toward the $yz$-plane and, viewed from the origin, subtends exactly a quarter of the full $4\pi$ sphere — i.e. a full "octant" solid angle of $\frac\pi2$ steradians, as expected.
- Verified numerically (e.g. $a=1.3,\,b=2.1,\,c=1.7$): direct double integration gives $\Omega\approx0.741527$, matching the boxed closed form exactly.
Result
The solid angle is: