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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

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Ok, so how did you get the 4 in front of the integral?

  4. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Thanks!!!

  7. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    \[\int\limits_{0}^{2\pi}\int\limits_{0}^{r}rdrd \Theta\] \[\int\limits_{0}^{2\pi}[r^2/2][0,r]d \Theta=\int\limits_{0}^{2\pi}r^2/2d \Theta\] \[\int\limits_{0}^{2\pi}r^2/2d \Theta=r^2/2(\Theta)|[0,2\pi]=\pi r^2\]

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