Find the area of the region that lies inside both curves:

r=3sin(2θ)
r=3sin(θ)

Question

Find the area of the region that lies inside both curves:

r=3sin(2θ)
r=3sin(θ)

anonymous · Answer

Find the point of intersection and use that as your boundary points when doing the integral.

anonymous · Answer

I did! Can you work it out and see what you get?

anonymous · Answer

$$\int\limits_{a}^{b} (1/2)r^2 dr$$ is the formula

anonymous · Answer

like i totally forgot how to graph in polr coordinates

anonymous · Answer

but i cant get it to work even trying different bounds

apoorvk · Answer

lol, i didnt read this was polar coordinates. i don't have much idea, but lets see.. i had drawn a complete graph, and was just about to post it stupidly, when i happened to see samjordon's comment above.

anonymous · Answer

Helppp meeee somoneeeee

anonymous · Answer

i think i am giving up cuz its 1 30 am and i didnt finish my own work and i will be up till 4 in the morning. sorry

anonymous · Answer

@KingGeorge  and @eliassaab can u guys come help out

kinggeorge · Answer

Well, let's find the intersection points.$$3\sin(\theta)=3\sin(2\theta)$$So $$\sin(\theta)=\sin(2\theta)$$The obvious solution is $\theta=0$. The other solutions are at $\theta=\pi/3, \;\;\pi, \;\;5\pi/3$

anonymous · Answer

We are solving for the area

anonymous · Answer

Remember it is in polar

kinggeorge · Answer

Those give us the bounds of integration. Give me a minute to make sure we're integrating the correct things here.

anonymous · Answer

ok

kinggeorge · Answer

After I've graphed it it looks a bit wonky. it looks sort of like |dw:1334641393785:dw|And we want to find the area of the shaded region.