# Area of parametric curve

1. Jun 26, 2012

### Sebobas

Hey Guys!

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

Find the area of the "loop" (I'm guessing it's called) formed by the set of parametrics:

--> x(t)=t3-3t and y(t)=t2+t+1

2. Relevant equations

I've already drawn the graph and as said before the curve creates a "loop", I have to find the area insed that "loop" and I don't know where to start.

3. The attempt at a solution

Here is the graph: http://www.wolframalpha.com/input/?i=Parametric+plot+(t^3-3t,t^2+t+1)

I know I probably have to integrate using y*dx..but I don't know where to separate the loop in order to integrate. I have no idea what the limits are.

Last edited: Jun 26, 2012
2. Jun 26, 2012

### SammyS

Staff Emeritus
Hello Sebobas. Welcome to PF !

Suppose that you know y as a function of x. In fact there would be one function for the upper portion of the "loop", call it f1(x), and another for the lower portion of the "loop", call it f2(x).

Let xL be the x value at the left-hand extreme of the "loop", and xR be the x value at the left-hand extreme of the "loop".

Then the area that you want is, of course,

$\displaystyle \large {\int_{x_L}^{x_R}}{f_1(x)}\,dx-\large {\int_{x_L}^{x_R}}{f_2(x)}\,dx\ .$

Also notice that $\displaystyle \int_{x_L}^{x_R}{f_2(x)}\,dx=-\int_{x_R}^{x_L}{f_2(x)}\,dx\ .$

And of course, as in integration by substitution, $\displaystyle \int f(x)\,dx=\int g(t)\,\frac{dx}{dt}dt\,,$ where g(t)=f(x(t)) .

3. Jun 26, 2012

### LCKurtz

That graph looks like the x-y coordinates of the intersection point are $(-2,3)$. You know $t$ is between -2.497 and 2.497. You can figure out the two values of $t$ by inspection.

4. Jun 26, 2012

### Sebobas

Hey Sammy! Thanks, glad to have found the community (:

I get what you mean, I even tried what you said, but I found myself with a problem..

...and that was that when trying to find y in function of x, t gave me 2 different values, which one do I use?

5. Jun 26, 2012

### Sebobas

Hey Kurtz.. well you are right the graphs do intersect at (2,3), when t = 1 and t =-2 ...but what do I do then? Separate the curves? in that case what are the upper and lower limits of those new integrals?

6. Jun 26, 2012

### LCKurtz

Well, you mentioned in your original post about integrating ydx as a line integral. What do you get if you integrate$\int_C -ydx + 0dy$ counterclockwise around a loop? Think about Green's theorem. And you can do the line integral as a function of $t$.

7. Jun 26, 2012

### SammyS

Staff Emeritus
You don't need to use y as a function of x. I was just showing how one might think about the problem.

Basically the g(t) I mentioned in that last step gives y as a function of t: y = f(1 OR 2)(x(t)) = t2 + t + 1 .

dx/dt = 3t2 - 3 .