## anonymous 5 years ago How do I use integration to prove that the area of a circle radius r is equal to pi r^2

1. Owlfred

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2. anonymous

The key is to use trigonometric substitution. If you start with: $4\int\limits_{0}^{r}\sqrt{r^2-x^{2}}dx$ Then make the following substitutions based on the angle:$x=r\cos \theta, \sqrt{r^2-x ^{2}}=r\sin \theta, dx=-r\sin \theta d\theta$ You'll get: $4\int\limits_{\pi/2}^{0}-r^2\sin^2 \theta d \theta$ or $4r^2\int\limits_{0}^{\pi/2}\sin^2 \theta d \theta$ I'll leave it to you to work out the definite integral.

3. anonymous

Ok, so how did you get the 4 in front of the integral?

4. anonymous

The integral represents the area of a quarter of a circle with radius r. I could have put 2 in front and worked out the sum from -r to r instead of 4 times 0 to r.

5. anonymous

Come to think of it, there's a much easier solution:$\int\limits_{0}^{r}2 \pi xdx$The integrand represents the circumference of a circle with radius x. The area of the circle is the cumulative sum of the circumference of the concentric circles from 0 to r.

6. anonymous

Thanks!!!

7. anonymous

Theres a simpler method If u divide the circle into many concentric rings, each at a distance r, and having thickness dr the area of one disc is $dA=2{\pi}rdr$ $\int\limits_{}^{}dA = \int\limits_{0}^{R}2{\pi}rdr = {\pi}R ^{2}$

8. anonymous

$\int\limits_{0}^{2\pi}\int\limits_{0}^{r}rdrd \Theta$ $\int\limits_{0}^{2\pi}[r^2/2][0,r]d \Theta=\int\limits_{0}^{2\pi}r^2/2d \Theta$ $\int\limits_{0}^{2\pi}r^2/2d \Theta=r^2/2(\Theta)|[0,2\pi]=\pi r^2$