Answer
For the parametrization $(t^3,t^6)$, the area is $\frac{1}{3}$.
For the parametrization $(t^2,t^4)$, the area is $\frac{1}{3}$.
Work Step by Step
1. Parametrization: $(t^3,t^6)$
So, we have
$x\left( t \right) = {t^3}$, $x'\left( t \right) = 3{t^2}$.
$y\left( t \right) = {t^6}$.
Since $x\left( t \right)$ is increasing, the $x$-interval $\left[ {0,1} \right]$ corresponds to the $t$-interval $\left[ {0,1} \right]$.
Using Eq. (11) the area is
$A = \mathop \smallint \limits_{t = 0}^1 {t^6}\left( {3{t^2}} \right){\rm{d}}t = 3\mathop \smallint \limits_{t = 0}^1 {t^8}{\rm{d}}t = \frac{3}{9}{t^9}|_0^1 = \frac{1}{3}$.
2. Parametrization: $(t^2,t^4)$
So, we have
$x\left( t \right) = {t^2}$, $x'\left( t \right) = 2t$.
$y\left( t \right) = {t^4}$.
Since $x\left( t \right)$ is increasing, the $x$-interval $\left[ {0,1} \right]$ corresponds to the $t$-interval $\left[ {0,1} \right]$.
Using Eq. (11) the area is
$A = \mathop \smallint \limits_{t = 0}^1 {t^4}\left( {2t} \right){\rm{d}}t = 2\mathop \smallint \limits_{t = 0}^1 {t^5}{\rm{d}}t = \frac{2}{6}{t^6}|_0^1 = \frac{1}{3}$.
