Parametric Intergration question

1. Jun 11, 2008

thomas49th

1. The problem statement, all variables and given/known data

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:

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

3. The attempt at a solution

Not sure what to do :\

2. Jun 11, 2008

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.

3. Jun 11, 2008

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?

4. Jun 11, 2008

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

5. Jun 11, 2008

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.

6. Jun 11, 2008

tiny-tim

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, $$\int y dx = \int y \frac{dx}{dt} \ dt$$

7. Jun 11, 2008

thomas49th

ahh yeh it's easy

got it!

cheers :)

8. Jun 12, 2008

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 :)