Similar to the original poster of this question Is this a valid proof for the area of a circle?, I am a high school AP Calc BC student using the idea of Riemann sums to add an infinite number of isosceles triangles of area
$\frac{R^{2}\sin(d\theta)}{2}$
to determine the area of the circle. This technique produces the sum:
$\lim_{n \to \infty } \sum_{i=1}^{n}\frac{R^{2}\sin(\frac{2\pi}{n})}{2}$
which does evaluate to $\pi R^{2}$. However, I would like to convert the sum to this integral: $\int_{0}^{2\pi}\frac{R^{2}\sin(d\theta)}{2}$, and then evaluate this intergal. Using nonstandard analysis, my guess is to simply $\sin(d\theta)$ as another infinitesimal dω, naively keeping the bounds the same and evaluating $\int_{0}^{2\pi}\frac{R^{2}}{2}d\omega$ does yield the wanted answer: $\pi R^{2}$? I fear this is an abuse of logic and/or notation and would like to know how to approach integrals like the one in question.
Is this a valid integral to prove area of circle
calculusintegrationnonstandard-analysis
Best Answer
That's a good intuition as $n \to \infty$ we can heuristically substitute $\sin (d\theta)$ with $d\theta$.
More rigourosly we have
$$\sum_{i=1}^{n}\frac{R^{2}\sin(\frac{2\pi}{n})}{2} = \sum_{i=1}^{n}\frac{R^{2}}{2}\frac{\sin(\frac{2\pi}{n})}{\frac{2\pi}{n}}\frac{2\pi}{n} $$
and since
$$\lim_{n \to \infty } \frac{\sin(\frac{2\pi}{n})}{\frac{2\pi}{n}} =1$$
we obtain
$$\lim_{n \to \infty } \frac{2\pi}{n}\sum_{i=1}^{n}\frac{R^{2}}{2}\frac{\sin(\frac{2\pi}{n})}{\frac{2\pi}{n}} = \int_{0}^{2\pi}\frac{R^{2}}{2} d\omega = \frac{R^{2}}{2}\int_{0}^{2\pi} d\omega =\frac{R^{2}}{2}\cdot 2\pi=\pi R^2$$
Refer also to: