# Area of parametric equation

1. Apr 14, 2005

### PolarClaw

Hi,

I've been trying to do this one question: Let R be the region enclosed by the graph x=t^2-2 y=t^3-2t. Set up the integral for the area of R.

I know that if y is continous function of x on an interval a ≤ x ≤ b where x=f(t) and y=g(t) then $$\int_{a}^{b}$$ y dx =$$\int_{t1}^{t2}$$ g(t)f'(t)dt provided that f(t1)=a and f(t2)=b and both g and f' are continous on [t1,t2].

But everytime i equate and solve for t i get sqrt(2) and 1. This is where i get lost I cant seem to find the interval for parametric equations (any of them ). Other then the interval I do belive i have the rest of the integral:

$$\int$$ (t^3-2t)(2t)dt

Thanks in advance (sorry for long post as well as horrid use of that latex typing first time)

2. Apr 14, 2005

### learningphysics

The graph translated to $$y=x\sqrt{x+2}$$. There doesn't seem to be any enclosed region.

3. Apr 14, 2005

### PolarClaw

the question copied and pasted from the pdf file is as follows:

4. Let R be the region enclosed by the graph of x = t^2 − 2, y = t^3 − 2t . (The graph was sketch for the final in 2004)

a. Set up, BUT DO NOT EVALUATE an integral for the area of R .

Thats all the info that was given :/

Last edited: Apr 14, 2005
4. Apr 14, 2005

### HallsofIvy

Staff Emeritus
Sorry, there is NO area "enclosed" by that graph.

5. Apr 14, 2005

### hypermorphism

Actually, the parametrization can be broken into the 3 implicit functions y = $x\sqrt{x+2}$ for $t\in(0, \infty)$, y = $-x\sqrt{x+2}$ for $t\in(-\infty, 0)$, and y = 0 for t = 0. The graphs of these functions enclose a region between x=0 and x=-2. Given the symmetry of the graphs of the functions (one is just a reflection in the x-axis of the other), all one needs to do is integrate the positive function on this interval and multiply by 2.

6. Apr 14, 2005

### PolarClaw

Thanks for the help got was able to get the answer now.

7. Apr 14, 2005

### learningphysics

Oops. Yes... the minus sign. Sorry about that.