Find area between 2 polar curves

  • Thread starter bavman
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  • #1
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Homework Statement



Fine the area of the region inside the polar curve r=4sin(theta) and outside the polar curve r=2.

Homework Equations



area under polar curve = 1/2 integral (a,b) r^2 d[tex]\Theta[/tex]

The Attempt at a Solution



I set up the integral like follows:

integral (0,pi/2) (4sin([tex]\Theta[/tex])^2 d[tex]\Theta[/tex] - integral (pi/2, pi/6) 2 d[tex]\Theta[/tex]

Then from plugging those into my calculator i got 4pi - (2/3)pi = (10/3) pi

Sorry i cant get the integral latex code to work for some reason.
 

Answers and Replies

  • #2
446
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integral (0,pi/2) (4sin([tex]\Theta[/tex])^2 d[tex]\Theta[/tex] - integral (pi/2, pi/6) 2 d[tex]\Theta[/tex]
Why do u think that the integral should be from [tex]0,\frac{\pi}{2}[/tex]?
Don't you think its the intersection(s) of the 2 polar curves?:grumpy:
 
  • #3
3
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hmm..well i really dont know how i messed that up.

so it should be :

integral (pi/6, 5pi/6) r1^2 - r2^2 dtheta ?
 
  • #4
446
1
That should be fine! :)
 

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