# Parametric question

1. Apr 3, 2005

### ILoveBaseball

Find the area of the region enclosed by the parametric equation
$$x=t^3-2t$$
$$y=9t^2$$

$$dx/dt = 3t^2-2$$
9t^2 - 1 = 0
$$t=\pm \sqrt {1/9}$$

$$\int_{-1/3}^{1/3} (9t^2)*(3t^2-2) dt$$

= -2/5

anyone know where i went wrong?

2. Apr 3, 2005

### quasar987

I don't understand what you did in line 3, 4, 5...

What I recommand doing is to find a explicit relation between x and y. This can be done in the following manner.

$$y = 9t^2 \Rightarrow t = \pm \sqrt{y/9}$$

In (1):

$$x(y) = \pm (y/9)^{3/2} \mp \frac{2}{3}\sqrt{y}$$

Now you got two curves. See where they intersect. See at which values of t does this corespond. Integrate acordingly (you did that right, except with the wrong bounds).

3. Apr 4, 2005

### ILoveBaseball

$$x(y) = \pm (y/9)^{3/2} \mp \frac{2}{3}\sqrt{y}$$

what am i suppose to do after this? solve for y? can you help me out a bit more? i'm just so confuse on how to find the bounds.

4. Apr 4, 2005

### quasar987

See where they intersect. Set them equal:

$$(y/9)^{3/2} - \frac{2}{3}\sqrt{y} = - (y/9)^{3/2} + \frac{2}{3}\sqrt{y}$$

and solve for y.

You'll find 2 y. Using $y=9t^2$, find to which values of t these two y corespond. These are your bounds.

5. Apr 4, 2005

setting those equal to each other and solving for y, i get $$\pm\sqrt{973}$$

those are the two y's right?

"Using y=9t^2, find to which values of t these two y corespond. These are your bounds"

are you saying to plug in $$\pm\sqrt{973}$$ into the y equation?

6. Apr 4, 2005

### quasar987

I don't get the same thing as you do. I get

$$(y/9)^{3/2} - \frac{2}{3}\sqrt{y} = - (y/9)^{3/2} + \frac{2}{3}\sqrt{y}$$

Right there, a solution is obviously y = 0. Let's find another.

$$2(y/9)^{3/2} = \frac{4}{3}\sqrt{y}$$

$$\frac{4}{729}y^3 = \frac{16}{9}y$$

$$\frac{4}{729}y^2 = \frac{16}{9}$$

$$y = \pm\sqrt{ \frac{16}{9}\frac{729}{4}} = \pm \sqrt{324} = \pm 18$$

This is a bit weird because $y=-18$ is not a selution because $\sqrt{-18}$ and $\sqrt{(-18)^3}$ are undefined. So y = +18 is our other solution. There are no other.

By "Using y=9t^2, find to which values of t these two y corespond. These are your bounds" I mean...

for which t will we have y=0 and for which t will be have y=18?

Well,

$$y= 0 = 9t^2 \Leftrightarrow t=0$$

and

$$y=18 = 9t^2 \Leftrightarrow t = \pm \sqrt{2}$$

We only need one. So let's discard the minus one.

Now integrate dxdy from 0 to sqrt{2} like you did before.