Answer
The area is $\frac{8~\sqrt{2}}{15}$
Work Step by Step
$y = \sqrt{t}$
$dy = \frac{1}{2\sqrt{t}}~dt$
Since the area is enclosed by the y-axis, we need to find the values of $t$ when $x=0$:
$x = t^2-2t = 0$
$t(t-2) = 0$
$t = 0, 2$
It would be be helpful to know if the area is on the left or the right of the y-axis.
When $t = 1$, $x = (1)^2-2(1) = -1 \lt 0$
Therefore, the area is to the left of the y-axis.
Thus, we should integrate from $2$ to $0$ to find the area.
We can find the area:
$A = \int_{2}^{0} x~dy$
$A = \int_{2}^{0} (t^2-2t)~\frac{1}{2\sqrt{t}}~dt$
$A = \frac{1}{2}\int_{2}^{0} (t^{3/2}-2\sqrt{t})~dt$
$A = \frac{1}{2}(\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2})\Big\vert_{2}^{0}$
$A = \frac{1}{2}[\frac{2}{5}(0)^{5/2}-\frac{4}{3}(0)^{3/2}]-\frac{1}{2}[\frac{2}{5}(2)^{5/2}-\frac{4}{3}(2)^{3/2}]$
$A = \frac{1}{2}[(0)-(\frac{8}{5}\sqrt{2}-\frac{8}{3}\sqrt{2})]$
$A = -\frac{1}{2}(\frac{24}{15}\sqrt{2}-\frac{40}{15}\sqrt{2})$
$A = -\frac{1}{2}(-\frac{16}{15}\sqrt{2})$
$A = \frac{8~\sqrt{2}}{15}$
The area is $\frac{8~\sqrt{2}}{15}$
