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

Is there any way to compute the integral $$\int_{-a}^{a} \sqrt{R^2-x^2} \,\mathrm{d}x, \qquad 0<a<R$$ 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.

• No that is not possible! Commented Dec 7, 2015 at 20:52
• If you are allowed to cheat and look up in a table of integrals… Commented Dec 7, 2015 at 21:00
• If you sketch the function under which you're integrating and the limits of integration you'll see that this is a part of the area of the circle with radius $R$. So if you wanted to do it geometrically, you could just use the formulas for the area of a (half) circle and for the area of a circular segment.
– user137731
Commented Dec 7, 2015 at 21:01
• The integral is 'half of the circle' - 'circular sector with $2\theta=2\cos^{-1}\frac aR$' + '2 triangles with base=$a$ and height=$\sqrt{R^2-a^2}$' Commented Dec 7, 2015 at 21:05
• Maybe you can adapt this: en.wikipedia.org/wiki/… Commented Dec 7, 2015 at 21:51

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}

• Wow, it's the first time I see such computation. Very nice work. Thank you.
– dh16
Commented Dec 8, 2015 at 3:34
• You're welcome. My pleasure. Commented Dec 8, 2015 at 3:37
• Wow. But how does one know to proceed like this without knowing what terms to add and subtract. These manipulations both scare and put me in awe. Commented Feb 4, 2021 at 13:02

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 $$\ln|\sec\theta+\tan\theta|+C$$ $$=R^2(\sec\theta\tan\theta|_{\theta=\tan^{-1}(-\dfrac{a}{R})}^{{\theta=\tan^{-1}(\dfrac{a}{R})}}+\ln|\sec\theta+\tan\theta|_{\theta=\tan^{-1}(-\dfrac{a}{R})}^{{\theta=\tan^{-1}(\dfrac{a}{R})}})$$

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.

• While useful information, this does not quite answer the question being asked! When posting an answer, please make sure that your post does indeed function as a standalone answer to the question. What you have posted is certainly a good note to provide, and perhaps could stand better as a comment rather than an answer! Commented May 26, 2022 at 4:26