TylerD
  • TylerD
Find the area of the region inside the rose r=4sin(2x) and inside the circle r=2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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TylerD
  • TylerD
x is theta, just used x for convenience, these are polar equations.
TylerD
  • TylerD
for reference https://www.desmos.com/calculator/icbkdjtg2n
TylerD
  • TylerD
the answer approx 9.82

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TylerD
  • TylerD
just having trouble finding that area there
TylerD
  • TylerD
i know the bounds for that area are 0 to 5pi/12
TylerD
  • TylerD
wait no i think i see now...
TylerD
  • TylerD
for the that rose petal, it should be from 0 to pi/2, but i gotta somehow subtract out the rest
TylerD
  • TylerD
gonna try something
ganeshie8
  • ganeshie8
try this 0->pi/12 : rose petal pi/12->pi/4 : circle
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=8*%28%5Cint%5Climits_0%5E%7Bpi%2F12%7D+1%2F2*%284*sin%282t%29%29%5E2+dt+%2B+%5Cint%5Climits_%7Bpi%2F12%7D%5E%7Bpi%2F4%7D+1%2F2*%282%29%5E2+dt%29
ganeshie8
  • ganeshie8
|dw:1438072090508:dw|
TylerD
  • TylerD
i think i see what u did just need to figure out how u factored it, sec.
ganeshie8
  • ganeshie8
that bottom "black" area is given by : \[\large \int\limits_0^{\pi/12} \frac{1}{2}*(4\sin(2t))^2 ~dt \]
ganeshie8
  • ganeshie8
the top blue area is given by : \[\large \int\limits_{\pi/12}^{\pi/4} \frac{1}{2}*(2)^2 ~dt \]
TylerD
  • TylerD
then you multiplied by 2 then again by 4? for symmetry
ganeshie8
  • ganeshie8
Yes
TylerD
  • TylerD
alright thanks, really appreciated, been stuck on that one for 2 hours.

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