finding the area between the parametric curve - Calc 2 help

[ltumechanicaltronics]

I need help finding the area between the parametric curve y=(t)^(1/2) x=(t^2)-2t and the Y axis. I found dx/dt=2t-2. I just guessed and checked in my ti-84 to find the interval.
I got S [0, 2pi/3] t^(1/2)*(t-2) (dt) which I integrated by parts to be (2/3)*(2t-2)*(t^(3/2))-(8/15)*(t^(5/2)) and plugged in 2pi/3 to get the answer
(8pi/9-4/3)(2pi/3)^(3/2)-(8/15)*(2pi/3)^(5/2). I've tried 3 different versions of this answer, and I have one last attempt to answer this problem. Is anything visibly wrong?

 

[skeeter]

where are the terms with [MATH]\pi[/MATH] coming from?

[MATH]y=\sqrt{t} \implies t \ge 0[/MATH]
[MATH]x=t^2-2t \implies x=y^4-2y^2[/MATH]
[MATH]x=0 \implies y=0 \text{ and } y= \sqrt{2}[/MATH]
looks like an integral of x as a function of y might be a bit easier to deal with ...

note the graphs of [MATH]\color{red}x=f(y)[/MATH] and [MATH]\color{blue}x=t^2-2t \text{ & } y=\sqrt{t}[/MATH]