Find circumference of the circle r=2acos theta.

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?

Sorry, slight typo. Please reasd it as Int (0 to 2pi)2a*Int d theta=2a(2pi-0)=4a*pi

well, 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.

Thanks a lot for guiding,please.

To find the circumference of a circle with radius r = 2acos(theta), the formula you provided is correct. However, there seems to be an error in your calculations.

Let's go through the steps again:

1. Start with the given integral:
s = ∫(0 to 2π) √(4a^2cos^2θ + 4a^2sin^2θ) dθ

2. Simplify the expression:
√(4a^2cos^2θ + 4a^2sin^2θ) = √(4a^2(cos^2θ + sin^2θ))
= √(4a^2)
= 2a

3. Substitute this back into the integral:
s = ∫(0 to 2π) 2a dθ

4. Integrate with respect to θ:
s = 2aθ ∣(0 to 2π)
s = 2a(2π - 0)
s = 4aπ

Therefore, the correct result is s = 4aπ, not 2aπ. It seems that there may have been a mistake or misunderstanding in the book.