I have actually a question around the area enclosed between the complying with parametric equations:

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\$.

You are watching: Find the area of the region enclosed by the parametric equation

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\$\$

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