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

This Question is Closed

TylerD
 one year ago
Best ResponseYou've already chosen the best response.1x is theta, just used x for convenience, these are polar equations.

TylerD
 one year ago
Best ResponseYou've already chosen the best response.1for reference https://www.desmos.com/calculator/icbkdjtg2n

TylerD
 one year ago
Best ResponseYou've already chosen the best response.1just having trouble finding that area there

TylerD
 one year ago
Best ResponseYou've already chosen the best response.1i know the bounds for that area are 0 to 5pi/12

TylerD
 one year ago
Best ResponseYou've already chosen the best response.1wait no i think i see now...

TylerD
 one year ago
Best ResponseYou've already chosen the best response.1for the that rose petal, it should be from 0 to pi/2, but i gotta somehow subtract out the rest

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2try this 0>pi/12 : rose petal pi/12>pi/4 : circle

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dw:1438072090508:dw

TylerD
 one year ago
Best ResponseYou've already chosen the best response.1i think i see what u did just need to figure out how u factored it, sec.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that bottom "black" area is given by : \[\large \int\limits_0^{\pi/12} \frac{1}{2}*(4\sin(2t))^2 ~dt \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2the top blue area is given by : \[\large \int\limits_{\pi/12}^{\pi/4} \frac{1}{2}*(2)^2 ~dt \]

TylerD
 one year ago
Best ResponseYou've already chosen the best response.1then you multiplied by 2 then again by 4? for symmetry

TylerD
 one year ago
Best ResponseYou've already chosen the best response.1alright thanks, really appreciated, been stuck on that one for 2 hours.
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