Area of between polar curves.

[klpierc2]

I am stuck on a problem in my Calc III class. Here is the problem: Find the area of the region between the two loops of the curve: r=1+2sinx. I have graphed it, and, after integrating, I am coming up with an area of around 10.5. I think my problem may be with the limits of integration. The limits I got on the outer curve were pi/2 to 11pi/6. and then subracting the inner curve with limits of 3pi/2 to 7pi/6. ( I used these limits b/c of the symmetry, I then doubled each. Am I using the correct limits? Thanks for any feedback!!

 

[galactus]

Someone will probably have a slicker method, but I will do this:

Entire area - small loop = area of region between large and small loops

\(\displaystyle \L\\\underbrace{\frac{1}{2}\int_{\frac{-\pi}{6}}^{\frac{7{\pi}}{6}}(1+2sin({\theta}))^{2}d{\theta}}_{\text{entire region:}\frac{3\sqrt{3}+4{\pi}}{2}}-\underbrace{\frac{1}{2}\int_{\frac{7{\pi}}{6}}^{\frac{11{\pi}}{6}}(1+2sin({\theta}))^{2}d{\theta}}_{\text{area small loop:}\frac{2{\pi}-3\sqrt{3}}{2}}\)

\(\displaystyle \L\\\frac{3\sqrt{3}+4{\pi}}{2}-\frac{2{\pi}-3\sqrt{3}}{2}=3\sqrt{3}+{\pi}\approx{\fbox{8.337745}}\)

See if anyone concurs. Check me out.

 

[soroban]

Hello, klpierc2!

Your limits are off . . . Is your sketch correct?

Find the area of the region between the two loops of the curve: \(\displaystyle \:r\:=\:1\,+\,2\sin x\)

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

My solution is the same as Galactus' . . .

\(\displaystyle \L\text{Area }\:=\:2\left[\frac{1}{2}\int^{\;\;\;\frac{\pi}{2}}_{-\frac{\pi}{6}} r^2\,d\theta \:-\:\frac{1}{2}\int^{\;\;\;\frac{3\pi}{2}}_{\frac{7\pi}{6}}r^2\,d\theta\right]\)