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Math Help - area of a circle segment

  1. #1
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    area of a circle segment



    is this suffice to calculate the area of a circle segment? Prove why or why not?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rmpatel5 View Post


    is this suffice to calculate the area of a circle segment? Prove why or why not?
    this integral gives the area "under" a straight line ax + b between the limits x_0 and x_1
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  3. #3
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    Hello, rmpatel5!

    Who asked this question ??
    It has nothing to do with a circle . . .


    \int^{x_1}_{x_o}(ax + b)\,dx

    Is this suffice to calculate the area of a circle segment? . . . . certainly not!
    Code:
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      - - * - - - - + - -a+:::*r- -
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    We have: . x^2+y^2 \:=\:r^2 \quad\Rightarrow\quad y \:=\:\sqrt{r^2 - x^2} . (the upper semicircle)

    The area is: . A \;=\;2 \times \int^r_a \left(r^2 - x^2\right)^{\frac{1}{2}}\,dx

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, rmpatel5!

    Who asked this question ??
    It has nothing to do with a circle . . .


    Code:
                    |
                  * * *
              *     |     *
            *       |     |:*
           *        |     |::*
                    |     |:::
          *         |     |:::*
      - - * - - - - + - -a+:::*r- -
          *         |     |:::*
                    |     |:::
           *        |     |::*
            *       |     |:*
              *     |     *
                  * * *
                    |

    We have: . x^2+y^2 \:=\:r^2 \quad\Rightarrow\quad y \:=\:\sqrt{r^2 - x^2} . (the upper semicircle)

    The area is: . A \;=\;2 \times \int^r_a \left(r^2 - x^2\right)^{\frac{1}{2}}\,dx

    hahah..my history of calculus teacher. He is kind of a nut. Also in your integral arnt u missing the part where u minus the integral of the line segment with the area of the circle??
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