Answer
(a) Using Eq. (11) the area is $\frac{2}{5}$.
(b) Using the standard method we get $y = {x^{3/2}}$. The area is also $\frac{2}{5}$.
Work Step by Step
(a) We have
$x\left( t \right) = {t^2}$, $x'\left( t \right) = 2t$,
$y\left( t \right) = {t^3}$.
Using Eq. (11) the area is
$A = \mathop \smallint \limits_{t = 0}^1 {t^3}\left( {2t} \right){\rm{d}}t = 2\mathop \smallint \limits_{t = 0}^1 {t^4}{\rm{d}}t = \frac{2}{5}{t^5}|_0^1 = \frac{2}{5}$.
(b) Since $x\left( t \right) = {t^2}$, we have $t = \pm \sqrt x $. Since $0 \le t \le 1$, we choose positive $t$. Substituting $t$ into $y$ gives $y = {x^{3/2}}$.
The $t$-interval $0 \le t \le 1$ corresponds to the $x$-interval $0 \le x \le 1$. So, using the standard method, the area under the graph $y = {x^{3/2}}$ is
$A = \mathop \smallint \limits_{x = 0}^1 {x^{3/2}}{\rm{d}}x = \frac{2}{5}{x^{5/2}}|_0^1 = \frac{2}{5}$.
