1. ## Parametric Equation

I have a question that I am working on and am having difficulties knowing how to approach it. It looks like this:

The following parametric equations trace out a loop.
x = 6 - (5/2)t^2
y = (-5/6)t^3+5t+1

A)Find the t values at which the curve intersects itself:
t= +/- ______
b) What is the total area inside the loop?

2. I know how to do PART A I am having difficulties with part B finding the area!

Part A:

-5/6 t³ + 5t + 1 = -5/6 [-t³] + 5[-t] + 1
-5/6 t³ + 5t = 0
t³ - 6t = 0
t = 0
t = +/-√6

For the Area

S = integral of y dx
S = integral of [-5/6 t³ + 5t + 1] d[6 - 5/2 t²]
S = integral of [-5/6 t³ + 5t + 1] 5tdt from T+ to T-
But I am not getting the right answer... Can anyone help?

3. Hello, SmileyFace143!

Just what is the right answer?
Give us something to aim at . . .

$t = \pm\sqrt{6}$

$A \;=\;\int^b_a y\,dx$

We have: . $A \;=\;\int^{\sqrt{6}}_{\text{-}\sqrt{6}}\left[-\frac{5}{6}t^3 + 5t + 1\right]\bigg[-5t\,dt\bigg] \;=\;\int^{\sqrt{6}}_{\text{-}\sqrt{6}}\left[\frac{25}{6}t^4 - 25t^2 - 5t\right]\,dt$

But I'm getting a negative answer . . .

4. There isnt a given answer. It is an online submission that just says right or wrong, with an unlimited number of tries. I have gotten the answer you came up with too -29 and it isnt correct so I am not too sure what I am doing wrong.