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I have a question about the area enclosed between the following parametric equations:

\begin{align*} x &= t^3 - 8t \\ y &= 6t^2 \end{align*}

I know the area is the integral of the $y(t)$ times the derivative of $x(t)$. What I don't know is how to find the limits of integration for $t$.

Thank you!

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    Do you mean the area bound between the curve & the x-axis (that is what you describe after the equations) ? There must be more in the question that you have not told us (yet) ? – 2017-02-20
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    It helps if you can sketch the graph to get an idea of what to integrate and what limits to use – 2017-02-20
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    @DavidQuinn Thanks David, right there is a loop ? they want the area enclosed in the loop ? – 2017-02-20
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    Yes there is a loop and they want the area enclosed in the loop – 2017-02-20

3 Answers 3

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by drawing a graph, e.g.

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

you can see that the loop is around points where $x = 0, y \ne 0$, that is $ t^3 - 8t = 0, t = +/- \sqrt8$, these are your limits, then as you said

$A = \int\limits_{-\sqrt8}^{\sqrt8} y(t) x'(t) dt = 1303.3...$

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    Narasimham , thank you! – 2017-02-20
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HINT

Use Green's thm between $t$ limits $\pm 2 \sqrt2$ that encloses a loop between the origin and $ y=48 $

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The graph has symmetry in the $y$ axis. The graph intersects with the $y$ axis when $t=0$ and $t=\pm2\sqrt{2}$

You therefore need to calculate $$A=2\int_{t=0}^{t=2\sqrt{2}}x\frac{dy}{dt}dt$$

Take the positive value of this.