area bounded by the curves...

**grgrsanjay** · June 7th 2011, 07:42 AM

Find The Area Bounded by the following curves
$x=a(\theta - \sin \theta$ )
$y=a(1-\cos \theta$
$0 \leqslant \theta \leqslant 2\pi$

THE LATEX IS AWESOME IN MHF NOW!!!

**Croat** · June 7th 2011, 09:17 AM

Well, I would first like to point out that what you have written aren't curves but just one curve defined parametrically. It's called a cycloid.
Anyway, I guess you'd like the area between it and the x-axis.
Just take the integral as always, but rewrite with theta:
$\int^{2a\pi}_0 y\,dx=\int^{2\pi}_0 a(1-\cos \theta)\,d(a(t-\sin \theta))=\int^{2\pi}_0 a^2(1-\cos \theta)(1-\cos \theta)\,dt$
$=\int^{2\pi}_0 a^2 (1-\cos \theta)^2\,dt$

Now you can solve this any way you like.

**TheEmptySet** · June 7th 2011, 09:19 AM

Originally Posted by grgrsanjay

Find The Area Bounded by the following curves
$x=a(\theta - \sin \theta$ )
$y=a(1-\cos \theta$
$0 \leqslant \theta \leqslant 2\pi$

THE LATEX IS AWESOME IN MHF NOW!!!

I assume you mean the area in the first quadrant

Just integrate

$\int_{0}^{2\pi}ydx=a^2\int_{0}^{2\pi}(1-\cos(\theta)^2dt$

**grgrsanjay** · June 7th 2011, 03:57 PM

Originally Posted by TheEmptySet

I assume you mean the area in the first quadrant

Just integrate

$\int_{0}^{2\pi}ydx=a^2\int_{0}^{2\pi}(1-\cos(\theta)^2dt$

i couldn't understand why are we doing it?

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