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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?

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    $\begingroup$ Not sure what's gained by the first integration by parts...just make the substitution $x=\sin \theta$ in the original integral. Also $\sin 1$ should never appear....I get that you introduce it and later take it out so the error causes no harm...but you really meant $\sin^{-1}(1)$ or $\frac {\pi}2$ and the lower limit should be $-\frac {\pi}2$. $\endgroup$ – lulu May 24 '16 at 17:45
  • $\begingroup$ Okay, so ignore the first integration by parts? And thanks for catching that limit error. I used symmetry to make lower bound $0$ there. $\endgroup$ – Lanier Freeman May 24 '16 at 17:48
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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.

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  • $\begingroup$ Great response, thanks. For a unit sphere, should I make $\sqrt{1-x^2}$ a rotational solid and integrate from there? $\endgroup$ – Lanier Freeman May 24 '16 at 19:55
  • $\begingroup$ No, staying in rectangular coordinates makes it easier to generalize to higher dimensions. What I am proposing is that you take a chainsaw to the unit sphere and slice through it parallel to the $yz$-plane. Can you find the radius, hence area of the circle formed by the cut at position $x$ along the $x$-axis? $\endgroup$ – user5713492 May 24 '16 at 20:11
  • $\begingroup$ Sure, it'd just be the unit circle of area $\pi$ at $x=0$, no? $\endgroup$ – Lanier Freeman May 24 '16 at 20:15
  • $\begingroup$ With radius approaching $0$ as $x$ approaches $\pm1$, too. $\endgroup$ – Lanier Freeman May 24 '16 at 20:21
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Yes,1.As the function is symmetric about $y$-axis too,hence integration can be called as twice the same integral from 0 to 1
2.Put $x=\sin(\theta)$ in first step.

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  • $\begingroup$ Would doubling the integrand and changing bounds have made calculations any easier? $\endgroup$ – Lanier Freeman May 24 '16 at 17:59
  • $\begingroup$ yes it helped (not much),but as complexity of question increases,it helps more $\endgroup$ – prashik May 24 '16 at 18:03
  • $\begingroup$ Alright, I"ll try to use that in the future whenever I can, thanks. $\endgroup$ – Lanier Freeman May 24 '16 at 18:04

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