polar coordinates area

**gracy** · December 19th 2006, 10:06 PM

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

**earboth** · December 20th 2006, 06:47 AM

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

**Soroban** · December 20th 2006, 06:49 AM

Hello, Gracy!

Find the area of the region that lies inside circle $r=1$
and outside the cardioid $r\:=\:1-\cos x$
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: . $1 \:= \:1 - \cos x$
. . Then: . $\cos x \,= \,0\quad\Rightarrow\quad x\:=\:-\frac{\pi}{2},\:\frac{\pi}{2}$

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

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

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

Go for it!

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