# 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 :)