MHB Find the area of the region which is inside both...

[shamieh]

Find the area of the region which is inside both$$ r = 2$$ and $$r = 4sin(\theta)$$How do I set up this? would I do..

$$\int ^{2\pi}_0 \frac{1}{2} [ r ] ^2 dr$$ ?

 

[soroban]

Hello, shamieh!

Obviously, you didn't make a sketch . . .

Find the area of the region which is inside both
r = 2 and r = 4\sin\theta

               4|
              * * *
          *     |     *
        *       |       *
       *        |        *
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      *        2|         *
      *       * * *       *
      *   *:::::|:::::*   *
        *:::::::|:::::::*
       *::::::::|::::::::*
        *:::::::|:::::::*
      *   *:::::|:::::*   *
  - - * - - - * * * - - - * - -
      *         |         *2
                |
       *        |        *
        *       |       *
          *     |     *
              * * *
                |

r = 2\,\text{ is a circle with center at }(0,0)\text{ and radius 2.}
r =4\sin\theta\,\text{ is a circle, center at }(0,2)\text{ and radius 2.}
. . [I'm speaking in rectangular coordinates, obviously.]

The circles intersect at \theta = \tfrac{\pi}{6},\:\tfrac{5\pi}{6}

Due to the symmetry, we can find the area of the right half
. . then multiply by 2.The area of the right half is:

. . \tfrac{1}{2}\int^{\frac{\pi}{6}}_0\!\! (4\sin\theta)^2d\theta \,+\, \tfrac{1}{2}\int^{\frac{\pi}{2}}_{\frac{\pi}{6}}\!\!(2^2)\,d\theta

Got it?

 

[shamieh]

YES! Awesome explanation!

 

[shamieh]

I ended up with $$A = -\sqrt{3}$$ now do I need to multiply it by 2 since it's symmetrical? to get $$A =-2\sqrt{3}$$?

 

[shamieh]

After setting up my integral properly I ended up with : $$A = \frac{8\pi}{3} + 2$$ is that what you got?