October 10th, 2017, 12:53 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 153 Thanks: 2  2D four leaves area problem
Find the area inside the fourleaf 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 07:08 PM. 
October 10th, 2017, 03:56 PM  #2 
Senior Member Joined: Jan 2017 From: Toronto Posts: 153 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, 04:08 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 1,664 Thanks: 844  
October 10th, 2017, 11:55 PM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 1,664 Thanks: 844  Quote:
$\displaystyle 4 \int_{\pi/6}^{\pi/6}\int_{1/2}^{\cos(2\theta)}~r~dr~d\theta$ The integral is one entire petal of the rose. There are 4 of them.  

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