Question

Show that the expression below is equal to the solid angle subtended by a rectangular aperture of sides $2a$ and $2b$ at a point a distance $c$ from the aperture along the normal to its centre: $$\mathrm{\Omega }=4{\int }_0^b\frac{ac}{(y^2+c^2){(y^2+c^2+a^2)}^{1/2}}dy$$ By setting $y={(a^2+c^2)}^{1/2}\tan \varphi$, change this integral into the form $${\int }_0^{{\varphi }_1}\frac{4ac\cos \varphi }{c^2+a^2{\sin }^2\varphi }d\varphi$$ where $\tan {\varphi }_1=b/{(a^2+c^2)}^{1/2}$, and hence show that $$\mathrm{\Omega }=4{\tan }^{-1}\left[\frac{ab}{c{(a^2+b^2+c^2)}^{1/2}}\right]$$

Short answer

$\mathrm{\Omega }=4{\tan }^{-1}\text{[}\frac{ab}{c\sqrt{a^2+b^2+c^2}}\text{]}$.

Step-by-step solution

Rewrite the integral using substitution

Use the substitution $y=\sqrt{a^2+c^2}\tan \varphi$. Calculate $dy$ and substitute into the integral. Remember $dy=\sqrt{a^2+c^2}(1+{\tan }^2\varphi )d\varphi$. Simplifying the expression leads to $dy=\sqrt{a^2+c^2}{\sec }^2\varphi d\varphi$.

Transform integral limits

Transform the limits of integration from y to $\varphi$. At $y=0$, $\varphi =0$; At $y=b$, $\varphi ={\varphi }_1$ where $\tan {\varphi }_1=\frac{b}{\sqrt{a^2+c^2}}$.

Transform integral function

Substitute $y$ and $dy$ into the integral $$4{\int }_0^b\frac{ac}{(y^2+c^2)(y^2+c^2+a^2{)}^{1/2}}dy$$ becomes $$4{\int }_0^{{\varphi }_1}\frac{ac\cos \varphi }{c^2+a^2{\sin }^2\varphi }d\varphi$$.

Use trigonometric identities to integrate

Simplify and use trigonometric identities to integrate the expression further. The integral can then be evaluated to give $$4{\tan }^{-1}\left[\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right]$$.

Key concepts

Rectangular Aperture

In this exercise, we need to find the solid angle subtended by a rectangular aperture. A rectangular aperture is simply a rectangular opening, often used in optical devices such as cameras and telescopes.
Its sides are defined as $2a$ and $2b$, forming a rectangle centered within the viewing field.
The distance from the center of this aperture to the observation point (or where we measure the solid angle) along the normal direction is denoted by $c$.
The main goal is to express the solid angle in a compact form, using specific mathematical substitutions and integrations.
A solid angle, much like a regular angle, is a measure of how big an object appears to an observer from a particular point.
It's measured in steradians, where the solid angle is a portion of the observer's sphere of vision.
For a rectangular aperture, the calculation involves understanding how the aperture's size and distance from the observer affect this spatial angle.

Integration by Substitution

Integration by substitution is a fundamental technique in calculus used to simplify integrals.
In this exercise, we substitute $y$ with $\sqrt{a^2+c^2}\tan \varphi$.
This change transforms our initial variable of integration into another that can make the integral easier to handle.
Substitution helps in breaking down complex expressions into manageable parts.
The differential $dy$ is recalculated as $\sqrt{a^2+c^2}{\sec }^2\varphi d\varphi$.
This new form aligns with trigonometric identities, paving the way to a simpler integral.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that simplify complex expressions.
For this problem, we use these identities to transform and simplify our integral.
One key identity is $1+{\tan }^2\varphi ={\sec }^2\varphi$, which helps in reshaping the integral function.
Another relevant identity is $\cos \varphi =\frac{1}{\sec \varphi }$, simplifying our work further.
These identities are essential because they allow us to manage the transformed integral efficiently.

Definite Integrals

A definite integral calculates the area under a curve between two limits.
In this case, our limits transform from $y$ to $\varphi$, representing the integration bounds.
Initially, the integral is from 0 to $b$ in the y-domain.
After substitution, the bounds change to 0 to ${\varphi }_1$, where $\tan {\varphi }_1=\frac{b}{\sqrt{a^2+c^2}}$.
Definite integrals not only simplify computations but provide a clear range within which to evaluate our functions.
This makes it easier to determine the exact solid angle.