When integrating under part of a circle, as in $$A=\int_0^a {\sqrt{r^2-x^2}\,\mathrm{d}x}$$ I noted that the simple geometric solution would be to add the areas of the sector and triangle formed by this shape (see picture below): $$A=[\text{sector}]+[\text{triangle}]=\left[{1\over 2}r^2\sin^{-1}{a\over r}\right]+\left[{1\over 2}a\sqrt{r^2-a^2}\right]$$

enter image description here

To perform the integration I would write $x=r\sin\theta,\;\mathrm{d}x=r\cos\theta\,\mathrm{d}\theta$ to get $$A=\int_0^{\sin^{-1}(a/r)}{r^2\cos^2\theta\;\mathrm{d}\theta}$$ Now using the double-angle formula, we have $$A={1\over 2}r^2\int_0^{\sin^{-1}(a/r)}{\big(1+\cos 2\theta\big)\;\mathrm{d}\theta}={r^2\over 2}\left[\theta+{1\over 2}\sin 2\theta\right]\Biggr|_0^{\sin^{-1}(a/r)}$$ and this is the same as before: $$A={r^2\over 2}\left[\sin^{-1}{a\over r}+{a\over r}\sqrt{1-{a^2\over r^2}}\;\,\right]$$

I was highly intrigued at how the substitution actually broke the definite integral into exactly the same pieces that I broke it into to use geometry. What is the meaning of this?

Also, on (perhaps) a related note, why is it that $$\int_0^{2\pi}{\sin^2 x\;\mathrm{d}x}=\pi$$ is the area of the unit circle? The fact that the parameter goes from $0$ to $2\pi$ makes me think this must have some simple area interpretation, but I can't find what it would be.

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    $\begingroup$ I'm not sure I got your second question right. But for a starters $\int_0^{2\pi} \sin^2 x dx = \int_0^{2\pi} \cos^2 x dx$, and latter is the same integral you wrote earlier but when $r = 1$. $\endgroup$ – Kaster Apr 24 '14 at 1:16

Since it is question of geometric interpretation of integration, we will come back three centuries ago in the time of G.W.Leibnitz and I.Newton. We will consider the integral as the sum of infinitesimal quantities, which are very small areas in the present case. The attached figure shows various manner to model the problem of the calculus of the area of a circle

enter image description here enter image description here

Of course, this manner to present intuitively a small value at limit reaching an infinitesimal $d\theta$ or $dx$ or $dy$ is a more "physical" than "mathematical" approach. This has been discussed a lot. More recently, the NonStandard Analysis provides a theoretical consistency. It is an important point, but so far from the scope of the present question !

Then, we can answer to the question initially raised. The figure below corresponds to the integral considered. While $x$ goes from $0$ to $a$, as shown on the figure, the surface scanned by the "hight part" is exactly the sector and the surface scaned by the "low part" is exactly the triangle.

enter image description here

  • $\begingroup$ A superb answer! I apologize for taking so long to accept it, I was confused because you wrote $x\cos\big(\theta_{(a)}\big)$ twice instead of $x\cot\big(\theta_{(a)}\big)$ (which is corrected everywhere after those initial steps) so I was trying to figure out the geometry. But I see now, thank you so much! A very insightful explanation :) $\endgroup$ – user142299 May 3 '14 at 20:37
  • $\begingroup$ I am sorry for the typo on the figure. All is well that ends well $\endgroup$ – JJacquelin May 7 '14 at 18:03
  • $\begingroup$ The typo has been corrected on the figure. Thank you to have brought the typing mistake to attention. $\endgroup$ – JJacquelin May 8 '14 at 7:23

Let $r(\theta)=2\sin(\theta)$ for $\theta\in[0,\pi]$. Then in polar coordinates this curve is a circle of radius $1$. Its area is $$\pi=\tfrac{1}{2}\int_0^{\pi}r^2(\theta)d\theta=\int_0^{2\pi}\sin^2(\theta)d\theta.$$

  • $\begingroup$ You should mention that this is not centered at the origin to avoid confusion. $\endgroup$ – Mikhail Katz Apr 27 '14 at 8:55
  • $\begingroup$ @WimC: it's better in the answer than in a comment that could be deleted. $\endgroup$ – robjohn Apr 27 '14 at 10:07

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