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December 18, 2014

December 18, 2014

Posted by **ms** on Monday, September 16, 2013 at 4:10am.

s= Int (0 to 2pi) of Sqrt(4a^2cos^2 theta+4a^2sin^2 theta)d theta

=Int (0 to 2pi)2a*Int theta d theta

=2a(2pi-0)=4a*pi

Book shows 2a*pi. Am I wrong somewhere?

- calculus -
**ms**, Monday, September 16, 2013 at 4:12amSorry, slight typo. Please reasd it as Int (0 to 2pi)2a*Int d theta=2a(2pi-0)=4a*pi

- calculus -
**Steve**, Monday, September 16, 2013 at 4:32amwell, you know from the equation that the circle has radius a, so its circumference is 2a*pi.

r = 2acosθ

r^2 = 2arcosθ

x^2+y^2 = 2ax

(x-a)^2 + y^2 = a^2

Now, as for the integration, your formula is correct, but as θ goes from 0 to 2pi, the circle is traced twice. So, you should only integrate from 0 to pi.

Do a plot and you can see why this is so.

- calculus -
**ms**, Monday, September 16, 2013 at 4:44amThanks a lot for guiding,please.

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