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Area of circle using integration

  1. Jan 9, 2016 #1
    1. The problem statement, all variables and given/known data
    http://postimage.org/] Screen_Shot_2016_01_09_at_10_13_38_AM.png [/PLAIN] [Broken]
    free picture upload


    2. The attempt at a solution
    I want to go width times delta height. To do this I must describe width in terms of height.

    Here they used the Pythagorean theorem which is weird to me because I don't see a nice triangle. Where h meets w the triangle extends outside of the circle. We divide w by 2 I guess to treat it as like a radius?

    I guess my question is how do I describe width in terms of height in this context?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 9, 2016 #2

    Mark44

    Staff: Mentor

    The triangle isn't shown in the picture. The vertical leg of the triangle goes from the center up to the horizontal strip. The horizontal leg is half the width of the strip. The hypotenuse is the radius of the circle.
    No. See above.
    They already give you w in terms of h in the equation ##w = 2\sqrt{3^2 - h^2}.##
     
    Last edited by a moderator: May 7, 2017
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