Please confirm area of cycloid: x=3(2?–sin 2?), y=3(1-cos2?)

[MAC-A-TAC]

Please confirm area of cycloid: x=3(2?–sin 2?), y=3(1-cos2?)

Hello.

I need someone to check my answer for the following problem:
Find the area under one arch of a cycloid described by the parametric equations,
x = 3(2? – sin 2?) and y = 3(1 – cos 2?). The limits for ? are ? and 0

What I have worked out is 18?. Is this correct. :?

As always, thank you for your help.

 

[galactus]

Re: Please confirm area of a cycloid

One thing to note is that the area of the cycloid is equal to 3 times the area of the circle that formed it.

\(\displaystyle 6\pi=2{\pi}r\)

\(\displaystyle r=3\)

Area of circle: \(\displaystyle {\pi}(3)^{2}=9{\pi}\)

3 times this is \(\displaystyle 27{\pi}\)

 

[BigGlenntheHeavy]

Re: Please confirm area of a cycloid

Another way:

Let x = f(t) = 3(2t-sin(2t)) and y = g(t) = 3(1-cos(2t)), 0<t<Pi

Then \(\displaystyle \int_{a}^{b}ydx\) = \(\displaystyle \int_{0}^{pi}g(t)f'(t)dt\) = 9\(\displaystyle \int_{0}^{pi}[1-cos(2t)][2-2cos(2t)]dt\) = 27pi.

 

[MAC-A-TAC]

Re: Please confirm area of a cycloid

Thank you for the reply and much easier way of solving.
Would you have time to take a look at my work shown in the attachment?
Where did I make my mistake?

Previous file wouldn't open so I changed the compressed jpg to a non-compressed GIF.

Thank you.