Parametric Intergration question

[thomas49th]

Homework Statement

a curve has parametric equations:
x = t - 2sin t
y = 1 - 2cos t


R is enclosed by the curve and the x axis. Show that the area of R is given by the integral:

\int^{\frac{5\pi}{3}}_{\frac{\pi}{3}} (1-2\cos t)^{2}

Homework Equations

The Attempt at a Solution

Not sure what to do :\

 

[CompuChip]

Suppose that the curve was given in the form y = f(x), then how would you calculate the area? Now do the variable substitution from x to t which is given.

 

[Defennder]

You might want to make use of this:
\int f(x) dx = \int f(t) \frac{dx}{dt} \ dt
By the way what's this mathematical property called?

 

[thomas49th]

ermm mathematical propety?

dx/dt = 1 - 2cost


i have to multiply this by f(t) (how do i found that) and then intergrate that with respect to t?

Thanks

 

[Defennder]

Yes, the integration is done wrt t. Isn't that given in the answer? And as for f(t), you should know that f(x) in the original integral f(x)dx represents a particular variable. That variable must now be expressed in terms of t for the integration to work.

 

[tiny-tim]

\int f(x) dx = \int f(t) \frac{dx}{dt} \ dt

Hi thomas49th!

I suspect you're confused by the (x) and the (t).

Just write it \int f dx = \int f \frac{dx}{dt} \ dt

or, in this case, \int y dx = \int y \frac{dx}{dt} \ dt

 

[thomas49th]

ahh yeh it's easy

got it!

cheers :)

 

[CompuChip]

The property is called substitution of variables. Probably you've seen it in this form:
Solve \int \sin(t) \cos(t) \, dt; let u = \sin(t), then du = \cos(t) \, dt so \int \sin(t) \cos(t) \, dt = \int u \, du = \frac12 u^2 + C = \frac12 \sin^2(t) + C.
You'll notice that in switching from the variable t to the variable u, I replaced \frac{du}{dt} dt by du.

If you have never seen this, forget it. Otherwise I hope it clarifies what just happened :)