Parametric equations and integration

[ARandomOWl]

I have a set of parametric equations given by:

[LATEX]x=t-2sin t[/LATEX]

[LATEX]y=1-2cos t[/LATEX]

[LATEX]0\leqslant t \leqslant 2 \pi[/LATEX]

When y=0: [LATEX]t=\frac{\pi}{3}, \frac{5 \pi}{3}[/LATEX]

The question: Show that the area enclosed by the curve and the x-axis is given by the integral:

[LATEX]\int_{\frac{\pi}{3}}^{\frac{5 \pi}{3}} (1-2cost)^2dt[/LATEX]

Can someone explain why the area is not given by:

[LATEX]\int_{\frac{\pi}{3}}^{\frac{5 \pi}{3}} (1-2cost)dt[/LATEX]

I noticed that:

[LATEX](1-2cost)^2 = y^2 = y \frac{dx}{dt}[/LATEX]

But I don't know of it's significance.
Thanks.

 

[Ratchit]

ARandomOWl,

Can someone explain why the area is not given by:

Sure, area = ∫y*dx . dx = d(t-2sin(t))*dt = (1-2cos(t))*dt . y = 1-2cos(t)

So, ∫y*dx = ∫(1-2cos(t))*(1-2cos(t))*dt = ∫(1-2cos(t))^2*dt

Ratch

 

[ARandomOWl]

Ah, I think I see it now. So the area is actually given by:

[LATEX]\int_{\frac{\pi}{3}}^{\frac{5 \pi}{3}} (1-2cost)dx[/LATEX]

Then I need to find [LATEX]\frac{dx}{dt}[/LATEX], rearrange for dx and substitute into the above. Got it, thanks

 

[MrAl]

Hi,

The dx becomes (dx/dt)*dt so you end up integrating with respect to t.

One caution here though, is that you cant just throw a solution at it and expect it to work every time. The catch is that if the curve goes below zero you'll end up with the net area instead of the total area. The best bet is to graph the function and see what you can spot, and evaluating the end points.