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Why area of circle is pi*r*r

  1. Aug 19, 2015 #1
    How can we find that using calculus
     
  2. jcsd
  3. Aug 19, 2015 #2
  4. Aug 19, 2015 #3

    DEvens

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    Calculus is the end product. But you need to see how to get there.

    Reaching WAAAAY back, I think to grade 9, or about age 14 for those not in the N. American system.

    Consider a circle. Now cut it up into quarters. Flip alternate quarters so you get a wiggly back-and-forth thing that has half the outside of the circle on one side, half on the other, and a radius on each end. Now cut the middle third out of each quarter and flip it. You still have half the outer edge of the circle on each side, and still a radius on each end. Keep doing that. Flip the middle third of each segment. What you are getting closer and closer to is a rectangle with long side equal to half the outside circumference of a circle, and short side equal to the radius. Each time we flip the middle we get closer and closer to this rectangle. And this wiggly shape always has the same area as a circle.

    The outside circumference is ##2 \pi r##. That's the definition of ##\pi##. Now we have a rectangle that has one side ##r## and the other side half of the circumference or ##\pi r##. So it's area is ##\pi r^2##.
     
  5. Aug 20, 2015 #4
    Much simpler than calculus... see how the greeks knew about it! Modern day interpretation of course...!
     
  6. Aug 20, 2015 #5
    There are lots of ways to get the results, and calculus is one of them. The calculation above is an application of limit.
     
    Last edited by a moderator: Aug 20, 2015
  7. Aug 23, 2015 #6
    You could also work in non Euclidean geometry and get other results.
     
  8. Aug 23, 2015 #7

    HallsofIvy

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    Since you ask specifically about Calculus, the equation of a circle of radius R, with center at the origin of the coordinate system, is [itex]x^2+ y^2= R^2[/itex]. We can simplify by looking only at the first quadrant where [itex]y= \sqrt{R^2- x^2}[/itex], x running from 0 to R. That will be 1/4 of the full circle so find that area and multiply by 4.

    The area of that quarter circle is given by [itex]\int_0^R y dx= \int_0^R (R^2- x^2)^{1/2} dx[/itex]. To do that integration, use the trig substitution [itex]x= sin(\theta)[/itex].
     
  9. Aug 24, 2015 #8
    Probably the simplest calculation with calculus is to use polar coordinates:

    [tex] A=\int_0^{2\pi}\int_0^R rdrd\theta=\int_0^{2\pi}\frac{1}{2}R^2d\theta=\pi R^2.[/tex]
     
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