anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ali2x2
  • 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
ali2x2
  • ali2x2
After I've graphed it it looks a bit wonky. it looks sort of like |dw:1439222172122:dw|
ali2x2
  • ali2x2
i got 4.1457477328045543537133819110877218

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ali2x2
  • ali2x2
nick?
anonymous
  • anonymous
yeah that what i got, but need the answer to part b, sorry i didn't specify that in the question
ali2x2
  • ali2x2
lol ok
anonymous
  • anonymous
i apologize
ali2x2
  • ali2x2
its ok
ali2x2
  • ali2x2
idk 2nd one, ask @ganeshie8
ali2x2
  • ali2x2
or google search
IrishBoy123
  • IrishBoy123
sounds glib but \(\pi .8^2\)
anonymous
  • anonymous
^dont understna @IrishBoy123 that not correct
zepdrix
  • zepdrix
Sorry was busy doing some stuff >.< So what are we trying to do? Find some area?
anonymous
  • anonymous
yes
IrishBoy123
  • 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)
zepdrix
  • 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! :)

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