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

[shamieh]

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

\(\displaystyle \int ^{2\pi}_0 \frac{1}{2} [ r ] ^2 dr\) ?

 

[soroban]

Hello, shamieh!

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

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

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$$r = 2\,\text{ is a circle with center at }(0,0)\text{ and radius 2.}$$
$$r =4\sin\theta\,\text{ is a circle, center at }(0,2)\text{ and radius 2.}$$
. . [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 \(\displaystyle A = -\sqrt{3}\) now do I need to multiply it by 2 since it's symmetrical? to get \(\displaystyle A =-2\sqrt{3}\)?

 

[shamieh]

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