I realize that I can consult integration tables for integrals like this, but I wanted to test my knowledge of integration techniques to solve this integral for the area of the unit circle.
$$2\int_{-1}^1\sqrt{1-x^2}dx$$
I've recently read up on some calculus II integration methods and wanted to see if my usage of those to find an antiderivative of this function is both logical and as quick as possible, so here we go.
$$2\int_{-1}^1\sqrt{1-x^2}dx$$
Using integration by parts:
$$=\left(2x\sqrt{1-x^2}\right)_{-1}^1+2\int_{-1}^1\frac{x^2}{\sqrt{1-x^2}}dx$$
Using trig substitution $x=\sin\theta$:
$$=\left(2x\sqrt{1-x^2}\right)_{-1}^1+4\int_0^{\pi/2}\frac{\sin^2\theta}{\sqrt{1-\sin^2\theta}}\cos\theta d\theta$$
Pythagorean identities:
$$=\left(2x\sqrt{1-x^2}\right)_{-1}^1+4\int_0^{\pi/2}{\sin^2\theta}d\theta$$
Using double angle formula for integrand:
$$=\left(2x\sqrt{1-x^2}\right)_{-1}^1+4\int_0^{\pi/2}\left[\frac{1}{2}-\frac{\cos2\theta}{2}\right]d\theta$$
Integrating:
$$=\left(2x\sqrt{1-x^2}\right)_{-1}^1+4\left[\frac{\theta}{2}-\frac{\sin2\theta}{4}\right]_0^{\pi/2}$$
Using double angle formula:
$$=\left(2x\sqrt{1-x^2}\right)_{-1}^1+\left[2\theta-2\sin\theta\cos\theta\right]_0^{\pi/2}$$
Using $\theta=\arcsin x$:
$$=\left(2x\sqrt{1-x^2}\right)_{-1}^1+\left[2\arcsin x-2x\sqrt{1-x^2}\right]_0^{1}$$
Manipulating bounds and solving:
$$=\left(2x\sqrt{1-x^2}\right)_{0}^1+\left[2\arcsin x\right]_0^{1}$$
$$=\pi$$
To reiterate, is the above a logical approach to finding the area of the unit circle via integration techniques, and can it be shortened considerably?
Best Answer
I am able to justify the integration by parts. $$\begin{align}A&=2\int_{-1}^1\sqrt{1-x^2}dx=\left.2x\sqrt{1-x^2}\right|_{-1}^1+2\int_{-1}^1\frac{x^2}{\sqrt{1-x^2}}\\ &=0+2\int_{-1}^1 \left[-\sqrt{1-x^2}+\frac1{\sqrt{1-x^2}}\right]dx=-A+2\int_{-1}^1\frac1{\sqrt{1-x^2}}dx\end{align}$$ Solving for the area, $$A=\int_{-1}^1\frac1{\sqrt{1-x^2}}dx=\int_{-\frac{\pi}2}^{\frac{\pi}2}\frac{\cos\theta\,d\theta}{\cos\theta}=\left.\theta\right|_{-\frac{\pi}2}^{\frac{\pi}2}=\pi$$ So your alternatives were to use the double angle formula or integration by parts and the latter approach actually worked out well. Now do a sphere! You know already the radius of the circle at which the plane $x=x_0$ cuts the sphere $x^2+y^2+z^2=1$. Your just computed that circle's area. Multiply by the thickness of a slice and integrate. How many dimensions can you do? In statistical mechanics, ordinarily we don't stop until Avogadro's number.