Question
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?
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?
Answers
ms
Sorry, slight typo. Please reasd it as Int (0 to 2pi)2a*Int d theta=2a(2pi-0)=4a*pi
Steve
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.
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.
ms
Thanks a lot for guiding,please.