Integral of $\sqrt{R^2-x^2}$

Question

Is there any way to compute the integral \begin{equation} \int_{-a}^{a} \sqrt{R^2-x^2} \,\mathrm{d}x, \qquad 0<a<R \end{equation} without using trig substitution or integration by parts?

I'm thinking to relate this to area of circle, but I couldn't find the relationship.

Thank you.

Answer 1

Without trigonometric substitution, integration by parts, or appeal to the geometric interpretation of the integral, we need to devise some quite fortuitous manipulations. We proceed, therefore, and write

$$\begin{align} \sqrt{R^2-x^2}&=\frac12 \left(\sqrt{R^2-x^2}+\sqrt{R^2-x^2}\right)\\\\ &=\frac12 \left(\sqrt{R^2-x^2}+\sqrt{R^2-x^2}\right)+\frac{x^2}{\sqrt{R^2-x^2}}-\frac{x^2}{\sqrt{R^2-x^2}}\\\\ =&\frac12 \left(\sqrt{R^2-x^2}-\frac{x^2}{\sqrt{R^2-x^2}}\right)+\frac12\left(\sqrt{R^2-x^2}+\frac{x^2}{\sqrt{R^2-x^2}}\right)\\\\ &=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12\left(R^2-x^2\right)\left(\frac{1}{\sqrt{R^2-x^2}}+\frac{x^2}{(R^2-x^2)^{3/2}}\right)\\\\ &=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12\left(R^2-x^2\right)\frac{d\left(x/\sqrt{R^2-x^2}\right)}{dx}\\\\ &=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12R^2\left(\frac{1}{1+\left(\frac{x}{\sqrt{R^2-x^2}}\right)^2}\right)\frac{d\left(x/\sqrt{R^2-x^2}\right)}{dx}\\\\ &=\frac{d}{dx}\left(\frac12 x\sqrt{R^2-x^2}+\frac12 R^2 \arctan\left(\frac{x}{\sqrt{R^2-x^2}}\right) \right) \end{align}$$

Therefore, we arrive at

$$\begin{align} \int_{-a}^a\sqrt{R^2-x^2}\,dx&=2\int_0^a \sqrt{R^2-x^2}\,dx\\\\ &=\left.\left( x\sqrt{R^2-x^2}+ R^2 \arctan\left(\frac{x}{\sqrt{R^2-x^2}}\right) \right)\right|_0^a\\\\ &= a\sqrt{R^2-a^2}+ R^2 \arctan\left(\frac{a}{\sqrt{R^2-a^2}}\right) \end{align}$$

Answer 2

If you want to use trigonometric identity at the last moment you will find $\sec^3\theta$ for what you have to integration by parts. Let's start :

$$\int_{-a}^{a}\sqrt{R^2-x^2} \ \mathrm dx$$ Let $x=R\tan\theta$ so $\mathrm dx=R\sec^2\theta \ \mathrm d\theta$ $$=\int_{x=-a}^{x=a}R\sqrt{1-\tan^2\theta}R\sec^2\theta \ \mathrm d\theta$$ $$=R^2\int_{x=-a}^{x=a}\sec^3\theta \ \mathrm d\theta$$ $$=R^2\int_{x=-a}^{x=a}\sec\theta\cdot\sec^2\theta \ \mathrm d\theta$$ If you do integration by parts then you will notice you will get recursive integration, I am just taking first part, here you get that $\sec\theta\tan\theta-\int\sec^3\theta\mathrm d\theta+\int \sec\theta\mathrm d\theta$ (ignoring $\sec^3\theta$ for simplicity) and integrand of $\sec\theta$ with respect to theta is $\ln|\sec\theta+\tan\theta|+C$ $$=R^2(\sec\theta\tan\theta|_{\theta=\tan^{-1}(-\dfrac{a}{R})}^{{\theta=\tan^{-1}(\dfrac{a}{R})}}+\ln|\sec\theta+\tan\theta|_{\theta=\tan^{-1}(-\dfrac{a}{R})}^{{\theta=\tan^{-1}(\dfrac{a}{R})}})$$

Answer 3

I can't help with the integration part but can sure help with your thought about relating this to a circle.

Think about a circle inside a sphere, if the radius of the sphere is R, the distance of the circle from the center of the sphere is x then;

Radius of the circle is $\sqrt{R^2 - x^2}$

You can do the integration that way too but, this doesn't help much.