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 October 23rd, 2012, 01:39 PM #1 Senior Member   Joined: Jun 2011 Posts: 164 Thanks: 0 Prove the area formula of a circle? Someone asked me this question: Given that there is a circle with radius r, prove that the area of the circle is pi * r^2 given that you know the circumference of a circle is 2pi*r and there are 2pi radians in one revolution. He asked me to do it using limits. I knew how to prove that using definite integral, but he says he wants me to do it using limits, and there are two ways to do it. I got stuck and come here for help I am not very good at Calculus yet, if there you could explain how you got to your answer when you tell me it, it would be great!
 October 23rd, 2012, 03:30 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs Re: Prove the area formula of a circle? The area of an n-gon is: $A_n=\frac{1}{2}nr^2\sin$$\frac{2\pi}{n}$$$ hence: $\lim_{n\to\infty}A_n=\frac{r^2}{2}\lim_{n\to\infty }$$\frac{2\pi\sin\(\frac{2\pi}{n}$$}{\frac{2\pi}{n }}\)=\pi r^2\lim_{n\to\infty}$$\frac{\sin\(\frac{2\pi}{n}$$ }{\frac{2\pi}{n}}\)=\pi r^2\cdot1=\pi r^2$
 October 23rd, 2012, 03:36 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond Re: Prove the area formula of a circle? You can divide a circle into several small triangles having two sides equal to the radius, r, and having angle ? opposite the arc of the circle. It's easier to deal with right angled triangles, so we bisect ? with a line down to the base of the triangle. The base is the segment between the two radii. There are 4?/? of these right angled triangles. Their area is $\sqrt{r^2\,-\,r^2\sin^2$$\frac{\theta}{2}$$}\,\cdot\,r\sin$$\f rac{\theta}{2}$$\,\cdot\,\frac12\,=\,r^2\sqrt{1\,-\,\sin^2$$\frac{\theta}{2}$$}\,\cdot\,\sin$$\frac{ \theta}{2}$$\,\cdot\,\frac12$ So our limit is $\lim_{\theta\to0}\,\frac{2\pi}{\theta}\,\cdot\,r^2 \sqrt{1\,-\,\sin^2$$\frac{\theta}{2}$$}\,\cdot\,\sin$$\frac{ \theta}{2}$$$ which can be written as $\lim_{\theta\to0}\,2\pi\,\cdot\,r^2\sqrt{1\,-\,\sin^2$$\frac{\theta}{2}$$}\,\cdot\,\frac{\frac1 2\sin$$\frac{\theta}{2}$$}{\frac{\theta}{2}}\,=\,2 \pi\,\cdot\,r^2\,\cdot\,1\,\cdot\,\frac12\,=\,\pi r^2$ I'm assuming you know $\lim_{x\to0}\,\frac{\sin(x)}{x}\,=\,1$
 October 23rd, 2012, 03:43 PM #4 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Prove the area formula of a circle? Actually when using integration you are using limit. $\int^{b}_{a} \,F(x)\, dx= \,\lim_{n\to \infty}\, \sum^{n}_{i=1} F(x_i)\,\Delta x\text{ where }x_i= a+i\Delta x\text{ and }\Delta x=\frac{b-a}{n}$
 October 23rd, 2012, 05:50 PM #5 Senior Member   Joined: Jun 2011 Posts: 164 Thanks: 0 Re: Prove the area formula of a circle? Thanks guys! That was very helpful, I did not realize circle could be cut into infinitely amount of pieces, which was probably why I struggled with it As for the integral, I don't think he ever wants to see an integral in this problem, so....

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