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TylerD

  • one year ago

Find the area of the region inside the rose r=4sin(2x) and inside the circle r=2

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  1. TylerD
    • one year ago
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    x is theta, just used x for convenience, these are polar equations.

  2. TylerD
    • one year ago
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    for reference https://www.desmos.com/calculator/icbkdjtg2n

  3. TylerD
    • one year ago
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    the answer approx 9.82

  4. TylerD
    • one year ago
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    just having trouble finding that area there

  5. TylerD
    • one year ago
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    i know the bounds for that area are 0 to 5pi/12

  6. TylerD
    • one year ago
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    wait no i think i see now...

  7. TylerD
    • one year ago
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    for the that rose petal, it should be from 0 to pi/2, but i gotta somehow subtract out the rest

  8. TylerD
    • one year ago
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    gonna try something

  9. ganeshie8
    • one year ago
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    try this 0->pi/12 : rose petal pi/12->pi/4 : circle

  10. ganeshie8
    • one year ago
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    |dw:1438072090508:dw|

  11. TylerD
    • one year ago
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    i think i see what u did just need to figure out how u factored it, sec.

  12. ganeshie8
    • one year ago
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    that bottom "black" area is given by : \[\large \int\limits_0^{\pi/12} \frac{1}{2}*(4\sin(2t))^2 ~dt \]

  13. ganeshie8
    • one year ago
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    the top blue area is given by : \[\large \int\limits_{\pi/12}^{\pi/4} \frac{1}{2}*(2)^2 ~dt \]

  14. TylerD
    • one year ago
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    then you multiplied by 2 then again by 4? for symmetry

  15. ganeshie8
    • one year ago
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    Yes

  16. TylerD
    • one year ago
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    alright thanks, really appreciated, been stuck on that one for 2 hours.

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