Answer
$A=\displaystyle \frac{8\sqrt{2}}{15}$
Work Step by Step
The y axis has equation $x=0.$
The curve intersects the y-axis when
$x=0$
$t^{2}-2t=0$
$t(t-2)=0$
... when $t=0$ and $t=2.$
For t=0$, y=0$, and for $t=2 ,\ y=\sqrt{2}$.
On the interval $y\in[0,\sqrt 2], (t\in[0,2])$, x(t) is negative, so we find
$A=\displaystyle \int_{0}^{\sqrt{2}}(-x(t))dy\qquad \left[\begin{array}{lll}
\frac{dy}{dt}=\frac{1}{2\sqrt{t}} & & y=0\Rightarrow t=0\\
dy=\frac{dt}{2\sqrt{t}} & & y=\sqrt{2}\Rightarrow t=2
\end{array}\right]$
$=-\displaystyle \int_{0}^{2}(t^{2}-2t)(\frac{1}{2\sqrt{t}}dt)$
$=-\displaystyle \int_{0}^{2}(\frac{1}{2}t^{3/2}-t^{1/2})dt$
$=-[\displaystyle \frac{1}{5}t^{5/2}-\frac{2}{3}t^{3/2}]_{0}^{2}$
$=-(\displaystyle \frac{1}{5}\cdot 2^{5/2}-\frac{2}{3}\cdot 2^{3/2})$
$=-\displaystyle \frac{\sqrt{2}}{15}(3\cdot 4-5\cdot 4)$
$=\displaystyle \frac{8\sqrt{2}}{15}$
