Answer
$\color{blue}{E(W) = \frac{1}{6}}$
Work Step by Step
$\underline{\text{First method:}}$ Use the pdf of $Y$.
\begin{align*}
E(W) &= E(Y^2) \qquad \text{since}\ W = Y^2 \\
&= \int_{\mathbb{R}} y^2\cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y^2)\ ]\\
&= \int_0^1 y^2\cdot(2(1-y))\ dy \qquad [\ \text{since}\ f_Y(y)=2(1-y),\ y\in[0,1]\ ] \\
&= \int_0^1(2y^2-2y^3)\ dy \\
&= \left( \frac{2}{3}y^3 - \frac{2}{4}y^4\right)_0^1 \\
&= \left( \frac{2}{3}1^3 - \frac{1}{2}1^4\right) - \left( \frac{2}{3}0^3 - \frac{1}{2}0^4\right) \\
&= \left(\frac{2}{3}-\frac{1}{2}\right) - (0-0) \\
\color{blue}{E(W)}\ &\color{blue}{= \frac{1}{6}}
\end{align*}
$\underline{\text{Second method:}}$ Use the pdf of $W$.
\begin{align*}
E(W) &= \int_{\mathbb{R}} w\cdot f_W(w)\ dw \qquad \left[\ \text{Definition of}\ E(W)\ \right]\\
&= \int_0^1 w\cdot\left( \frac{1}{\sqrt{w}}-1\right)\ dy \quad \left[\ \text{since}\ f_W(w)= \frac{1}{\sqrt{w}}-1,\ w\in[0,1]\ \right] \\
&= \int_0^1 \left(w^{1/2}-w\right)\ dy \\
&= \left( \frac{w^{3/2}}{3/2} - \frac{w^2}{2}\right)_0^1 \\
&= \left( \frac{2}{3}1^{3/2} - \frac{1}{2}1^2\right) - \left( \frac{2}{3}0^{3/2} - \frac{1}{2}0^2\right) \\
&= \left(\frac{2}{3}-\frac{1}{2}\right) - (0-0) \\
\color{blue}{E(W)}\ &\color{blue}{= \frac{1}{6}}
\end{align*}
