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September 16, 2014

September 16, 2014

Posted by **Log** on Saturday, March 21, 2009 at 8:17pm.

r=5+2sin(theta)

- Calculus -
**Count Iblis**, Saturday, March 21, 2009 at 9:39pmYou have to integrate 1/2 r^2 from theta = 0 to 2 pi.

1/2 r^2 = 1/2 [25 + 20 sin(theta)

+ 4 sin^2(theta)]

The sin(theta) term doesn't contribute to the integral and sin^2(theta) can be replaced by 1/2 [because it would yield the same contribution as cos^2(theta) and together they are equal to 1].

So, the area is given by:

pi[25 + 2 ] = 27 pi

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