My Math Forum 2D four leaves area problem

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 October 10th, 2017, 11:53 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 188 Thanks: 2 2D four leaves area problem Find the area inside the four-leaf rose $\displaystyle r = \cos( 2 \theta )$ and outside $r$ = 1/2. Answer: $\displaystyle \sqrt{3} / 4 + \pi / 6$ My solution: $\displaystyle \int_{0}^{ \pi /4 } \int_{1/2}^{ \cos {2 \theta} } 8 ~r~dr~d \theta = \pi /4$ What did I do wrong? Last edited by skipjack; October 10th, 2017 at 06:08 PM.
 October 10th, 2017, 02:56 PM #2 Senior Member   Joined: Jan 2017 From: Toronto Posts: 188 Thanks: 2 I got it!!! $\displaystyle \int_{0}^{ \pi /6 } \int_{0}^{ \cos {2 \theta } } 8 ~r~dr~d \theta - \int_{0}^{ \pi / 6 } \int_{0}^{1/2} 8 ~r~dr~d \theta = \sqrt{3} / 4 + \pi / 6$
 October 10th, 2017, 03:08 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,093 Thanks: 1087
October 10th, 2017, 10:55 PM   #4
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Quote:
 Originally Posted by zollen I got it!!! $\displaystyle \int_{0}^{ \pi /6 } \int_{0}^{ \cos {2 \theta } } 8 ~r~dr~d \theta - \int_{0}^{ \pi / 6 } \int_{0}^{1/2} 8 ~r~dr~d \theta = \sqrt{3} / 4 + \pi / 6$
there is a clearer way of writing this

$\displaystyle 4 \int_{-\pi/6}^{\pi/6}\int_{1/2}^{\cos(2\theta)}~r~dr~d\theta$

The integral is one entire petal of the rose. There are 4 of them.

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