## anonymous one year ago Find the area of the region that lies inside both curves r=3sin(2θ),r=3sin(θ) b. Find the area of the region bounded by: r=8−2sinθ

1. ali2x2

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

2. ali2x2

After I've graphed it it looks a bit wonky. it looks sort of like |dw:1439222172122:dw|

3. ali2x2

i got 4.1457477328045543537133819110877218

4. ali2x2

nick?

5. anonymous

yeah that what i got, but need the answer to part b, sorry i didn't specify that in the question

6. ali2x2

lol ok

7. anonymous

i apologize

8. ali2x2

its ok

9. ali2x2

idk 2nd one, ask @ganeshie8

10. ali2x2

11. IrishBoy123

sounds glib but $$\pi .8^2$$

12. anonymous

^dont understna @IrishBoy123 that not correct

13. zepdrix

Sorry was busy doing some stuff >.< So what are we trying to do? Find some area?

14. anonymous

yes

15. IrishBoy123

sorry @nick1234567 foolishly thought it was just a circle doing integration $$0 \rightarrow 2\pi$$ yields $$66 \pi$$ which sounds right as it is very close to be ing a circle origin (0,-2)

16. zepdrix

$\large\rm A=\int\limits_{\theta=\alpha}^{\beta}\frac{1}{2}r^2 d\theta$ $\large\rm A=\int\limits_{0}^{2\pi}\frac{1}{2}(8-2\sin \theta)^2d \theta$Do you understand setting up the integral? Maybe a few tricky steps from there, shouldn't be too bad though. I'm ending up with 66pi as well! :)