# polar coordinates area

• Dec 19th 2006, 10:06 PM
gracy
polar coordinates area
Find the area of the region that lies inside circle R=1 and out side the cardioid r=1-cos(x) sketch the graph
• Dec 20th 2006, 06:47 AM
earboth
Quote:

Originally Posted by gracy
Find the area of the region that lies inside circle R=1 and out side the cardioid r=1-cos(x) sketch the graph

Hello Gracy,

I'm not quite certain which area you mean.

I've attached a drawing of the circle and the cardioide where I've greyed the region which I believe should be calculated. Am I right?

EB
• Dec 20th 2006, 06:49 AM
Soroban
Hello, Gracy!

Quote:

Find the area of the region that lies inside circle $\displaystyle r=1$
and outside the cardioid $\displaystyle r\:=\:1-\cos x$
Sketch the graph.

Code:

                          |                   o  o  |               o        * o *             o      *    |  o::*           o      *      |    o::*         o      *        |    o:::*                           |    o:::::         o      *        | -o::::::*     ----o-------*---------o:::::::::*----         o      *        |  o::::::*                           |    o:::::         o      *        |    o:::*           o      *      |    o::*             o      *    |  o  *               o        * o *                   o  o  |                           |

The curves intersect when: .$\displaystyle 1 \:= \:1 - \cos x$
. . Then: .$\displaystyle \cos x \,= \,0\quad\Rightarrow\quad x\:=\:-\frac{\pi}{2},\:\frac{\pi}{2}$

Due to the symmetry, we can integrate from $\displaystyle 0$ to $\displaystyle \frac{\pi}{2}$ and double.

The shaded area is: .(Area of the circle) - (Area of the cardiod)

Therefore: .$\displaystyle A \;=\;2 \times \frac{1}{2}\int^{\frac{\pi}{2}}_0\left[1^2 - (1 - \cos x)^2\right]\,dx$

Go for it!