1
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ 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) ? $\endgroup$ Feb 20 '17 at 21:25
  • $\begingroup$ It helps if you can sketch the graph to get an idea of what to integrate and what limits to use $\endgroup$ Feb 20 '17 at 21:30
  • $\begingroup$ @DavidQuinn Thanks David, right there is a loop ? they want the area enclosed in the loop ? $\endgroup$ Feb 20 '17 at 21:37
  • $\begingroup$ Yes there is a loop and they want the area enclosed in the loop $\endgroup$ Feb 20 '17 at 21:40
1
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ Narasimham , thank you! $\endgroup$
    – Alex
    Feb 20 '17 at 22:16
0
$\begingroup$

HINT

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.