Find the area bounded with the curve $x=1+t-t^3, \ \ y=1-15t^2.$

Question

Find the area bounded with the curve $$x=1+t-t^3, \ \ y=1-15t^2.$$

I drew its graph and looks like this:

It seems to be symmetric about $x=1$. So, the idea that came to me was to cut by half and study any of the pieces. However, I don't know how... Any help is appreciated.

Answer 1

If $1+t-t^3=1+u-u^3,\,1-15t^2=1-15u^2,\,t\ne u$ then$$t\ne0,\,u=-t,\,1+t-t^3=1-t+t^3\implies 2t(1-t^2)=0\implies t=\pm1.$$So the area is from $t=-1$ to $t=1$. As $t$ increases from $-1$ to $0$, we trace out the left-hand part of the area's circumference; from $t=0$ to $t=1$, we follow the right-hand part. The shape vertically stretches from $t=\pm1,\,y=-14$ to $t=0,\,y=1$. From $y$ to $y+dy=y-30tdt$, a strip exists of width $2t(1-t^2)$, so the area is$$\int_0^160t^2(1-t^2)dt=60\int_0^1(t^2-t^4)dt=8.$$As a sanity check, the area fits in a rectangle of height $15$ and width $4/(3\sqrt{3})$ (because $x$ is extremised at $t=\pm1/\sqrt{3},\,x=1\mp2/(3\sqrt{3})$), so of area $20/\sqrt{3}>8$.

Answer 2

Cut the curve into left and right values for $x$ at each $y$ and make use of the symmetry. $$ {{\int_{-14}^{1} x_r-x_\ell\,\mathrm d y }={2\int_0^1 (x(t)-1)\,\left\lvert\frac{\mathrm d y(t)}{\mathrm d t}\right\rvert\,\mathrm d t}}$$