# Math Help - area of a circle segment

1. ## area of a circle segment

is this suffice to calculate the area of a circle segment? Prove why or why not?

2. 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 $ax + b$ between the limits $x_0$ and $x_1$

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

4. Originally Posted by Soroban
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??