# area of a circle segment

• Sep 12th 2008, 01:58 PM
rmpatel5
area of a circle segment

is this suffice to calculate the area of a circle segment? Prove why or why not?
• Sep 12th 2008, 02:22 PM
Jhevon
Quote:

Originally Posted by rmpatel5

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 $\displaystyle ax + b$ between the limits $\displaystyle x_0$ and $\displaystyle x_1$
• Sep 12th 2008, 02:23 PM
Soroban
Hello, rmpatel5!

It has nothing to do with a circle . . .

Quote:

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

Is this suffice to calculate the area of a circle segment? . . . . certainly not!

Code:

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

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

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

• Sep 12th 2008, 02:53 PM
rmpatel5
Quote:

Originally Posted by Soroban
Hello, rmpatel5!

                |               * * *           *    |    *         *      |    |:*       *        |    |::*                 |    |:::       *        |    |:::*   - - * - - - - + - -a+:::*r- -       *        |    |:::*                 |    |:::       *        |    |::*         *      |    |:*           *    |    *               * * *                 |
We have: .$\displaystyle x^2+y^2 \:=\:r^2 \quad\Rightarrow\quad y \:=\:\sqrt{r^2 - x^2}$ . (the upper semicircle)
The area is: .$\displaystyle A \;=\;2 \times \int^r_a \left(r^2 - x^2\right)^{\frac{1}{2}}\,dx$