eginalign* x &= t^3 - 8t \ y &= 6t^2endalign*

I understand the area is the integral that the $y(t)$ time the derivative of $x(t)$. What ns don"t know is just how to find the boundaries of integration because that $t$.

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Thank you!

by drawing a graph, e.g.

http://www.wolframalpha.com/input/?i=draw+x+%3D+t%5E3-8t,++y+%3D+6t%5E2

you deserve to see the the loop is about points where $x = 0, y e 0$, that is $ t^3 - 8t = 0, t = +/- sqrt8$, these space your limits, then together you said

$A = intlimits_-sqrt8^sqrt8 y(t) x"(t) dt = 1303.3...$

HINT

Use Green"s thm in between $t$ borders $pm 2 sqrt2$ the encloses a loop between the origin and $ y=48 $

The graph has symmetry in the $y$ axis. The graph intersects v the $y$ axis when $t=0$ and also $t=pm2sqrt2$

You therefore need to calculation $$A=2int_t=0^t=2sqrt2xfracdydtdt$$

Take the hopeful value that this.

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