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
by drawing a graph, e.g.
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
$A = intlimits_-sqrt8^sqrt8 y(t) x"(t) dt = 1303.3...$
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.
Thanks for contributing response to tastecraftedmcd.comematics Stack Exchange!Please be certain to answer the question. Administer details and also share your research!
But avoid …Asking because that help, clarification, or responding to various other answers.Making statements based on opinion; back them up with recommendations or personal experience.
Use tastecraftedmcd.comJax to layout equations. tastecraftedmcd.comJax reference.
See more: Why Is Dot Hack Quarantine So Expensive ? Hack Quarantine Prices Playstation 2
To find out more, view our advice on writing great answers.
short article Your price Discard
Not the prize you're spring for? Browse various other questions tagged calculus or ask your own question.
site design / logo design © 2021 stack Exchange Inc; user contributions licensed under cc by-sa. Rev2021.12.22.41046