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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
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

This Question is Closed

ali2x2
 one year ago
Best ResponseYou've already chosen the best response.1Well, 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

ali2x2
 one year ago
Best ResponseYou've already chosen the best response.1After I've graphed it it looks a bit wonky. it looks sort of like dw:1439222172122:dw

ali2x2
 one year ago
Best ResponseYou've already chosen the best response.1i got 4.1457477328045543537133819110877218

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah that what i got, but need the answer to part b, sorry i didn't specify that in the question

ali2x2
 one year ago
Best ResponseYou've already chosen the best response.1idk 2nd one, ask @ganeshie8

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1sounds glib but \(\pi .8^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0^dont understna @IrishBoy123 that not correct

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Sorry was busy doing some stuff >.< So what are we trying to do? Find some area?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1sorry @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)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm A=\int\limits_{\theta=\alpha}^{\beta}\frac{1}{2}r^2 d\theta\] \[\large\rm A=\int\limits_{0}^{2\pi}\frac{1}{2}(82\sin \theta)^2d \theta\]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! :)
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