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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

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  1. ali2x2
    • one year ago
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    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
    • one year ago
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    After I've graphed it it looks a bit wonky. it looks sort of like |dw:1439222172122:dw|

  3. ali2x2
    • one year ago
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    i got 4.1457477328045543537133819110877218

  4. ali2x2
    • one year ago
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    nick?

  5. anonymous
    • one year ago
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    yeah that what i got, but need the answer to part b, sorry i didn't specify that in the question

  6. ali2x2
    • one year ago
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    lol ok

  7. anonymous
    • one year ago
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    i apologize

  8. ali2x2
    • one year ago
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    its ok

  9. ali2x2
    • one year ago
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    idk 2nd one, ask @ganeshie8

  10. ali2x2
    • one year ago
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    or google search

  11. IrishBoy123
    • one year ago
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    sounds glib but \(\pi .8^2\)

  12. anonymous
    • one year ago
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    ^dont understna @IrishBoy123 that not correct

  13. zepdrix
    • one year ago
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    Sorry was busy doing some stuff >.< So what are we trying to do? Find some area?

  14. anonymous
    • one year ago
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    yes

  15. IrishBoy123
    • one year ago
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    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
    • one year ago
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    \[\large\rm A=\int\limits_{\theta=\alpha}^{\beta}\frac{1}{2}r^2 d\theta\] \[\large\rm A=\int\limits_{0}^{2\pi}\frac{1}{2}(8-2\sin \theta)^2d \theta\]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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