Question

Find the area enclosed by the curve x = t2 − 2t, y = t and the y-axis.

Answer 1

The curve hits the y-axis whenever $x=0$:

$x=t^2-2t=0\implies t(t-2)=0\implies t=0,t=2$

The area enclosed by the region is the sum of distances of every point on the curve where $t\in[0,2]$ to the origin. This is given by the integral

$\displaystyle\int_0^2\sqrt{x(t)^2+y(t)^2}\,\mathrm dt$
$\displaystyle\int_0^2\sqrt{(t^2-2t)^2+t^2}\,\mathrm dt$
$\displaystyle\int_0^2\sqrt{t^2(t-2)^2}\,\mathrm dt$
$\displaystyle\int_0^2|t||t-2|\,\mathrm dt$

Since $t\in[0,2]$, you have $|t|=t$ and $|t-2|=-(t-2)=2-t$, giving you

$\displaystyle\int_0^2(2t-t^2)\,\mathrm dt=t^2-\dfrac{t^3}3\bigg|_{t=0}^{t=2}=4-\dfrac83=\dfrac43$