# Math Help - area with polar coodinates

1. ## area with polar coodinates

I need the area inside r=18sin(theta) but outside r=2

Im having trouble finding the interval on which to integrate.

2. Hello, casesam!

Find the area inside $r\:=\:18\sin\theta$, but outside $r\:=\:2$
A sketch might help . . .
Code:
              18|
* * *
*:::::|:::::*
*:::::::|:::::::*
*::::::::|::::::::*
:::::::::|:::::::::
*:::::::::|:::::::::*
*:::::::::|:::::::::*
*:::::::::|:::::::::*
:::::::ooo:::::::
*:::o    |    o:::*
*:o     |     o:*
*     |     *
-----o----*-*-*----o-----
|      2
o     |     o
o   |   o
ooo
|

The graphs intersect where: . $18\sin\theta \:=\:2 \quad\Rightarrow\quad \sin\theta \:=\:\tfrac{1}{9}$

Hence: . $\theta \;=\;\sin^{-1}(\tfrac{1}{9}) \;\approx\;\begin{Bmatrix}0.11134 \\ 3.03025 \end{Bmatrix}$

Due to the symmetry, we can integrate from 0.11134 to $\tfrac{\pi}{2}$ . . . and double.

3. Thanks for the graph! so now for the integration

(2)(1/2)(integral from 0.11134 to (pi/2)) (18sin(theta))^2-(2)^2

=(1)(integral) 324(sin(theta))^2-4
=(4)(integral) 81(sin(theta))^2-1
=(4)(integral)81((1/2)(1-cos(2theta)))-1
=(4)(integral) (81/2)-(81/2)cos(2theta)-1
=4[(79/2)theta-(81/4)sin(2theta)] integrated from 0.11134 to (pi/2)

Is this right? and what final answer do you get with many decimal places please.