Monday

October 20, 2014

October 20, 2014

Posted by **Bobby** on Wednesday, March 10, 2010 at 9:33am.

- calculus -
**drwls**, Wednesday, March 10, 2010 at 10:29amUse a polar coordinate integration to get the area inside the rose. The circumscribed circle area is, of course, pi, since rmax = 1

For the rose, the area is

S (1/2)r^2 dx

(for x integrated from 0 to 2 pi

= S (1/2)sin^2x dx

"S" denotes an integral sign.

=(1/2)[(x/2) -(1/4)(sin2x)]@x=2pi -

(1/2)[(x/2) -(1/4)(sin2x)]@x=0

= (1/2)*pi

qed

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