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To set up the integral for $I_y$ using a vertical strip, first express the differential area $dA$ in terms of $x$. A vertical strip has an infinitesimal width $dx$ and a height determined by the difference between the upper and lower boundary curves at position $x$.

Conditions: The region is bounded by curves that can be expressed as functions of $x$.; A vertical differential strip is chosen, meaning its width is $dx$.; The integration variable is $x$.

To set up the integral for $I_y$ using a vertical strip, first express the differential area $dA$ in terms of $x$. A vertical strip has an infinitesimal width $dx$ and a height determined by the difference between the upper and lower boundary curves at position $x$.

Conditions: The region is bounded by curves that can be expressed as functions of $x$.; A vertical differential strip is chosen, meaning its width is $dx$.; The integration variable is $x$.

The final numerical value of the area moment of inertia $I_y$ for the specific shaded region about the centroidal y-axis is approximately $0.762 \text{ m}^4$. This value is obtained by evaluating the definite integral $\int_0^2 x^2 (2 - \sqrt{2} x^{1/2}) dx$, which simplifies to $\frac{2^4}{3} - \sqrt{2} \left(\frac{2}{7}\right) 2^{3.5}$.

Conditions: The region is bounded by $y = 2$ (top) and $y = \sqrt{2} x^{1/2}$ (bottom) from $x=0$ to $x=2$.; The axis is the centroidal y-axis.; Units are in meters.

The final numerical value of the area moment of inertia $I_y$ for the specific shaded region about the centroidal y-axis is approximately $0.762 \text{ m}^4$. This value is obtained by evaluating the definite integral $\int_0^2 x^2 (2 - \sqrt{2} x^{1/2}) dx$, which simplifies to $\frac{2^4}{3} - \sqrt{2} \left(\frac{2}{7}\right) 2^{3.5}$.

Conditions: The region is bounded by $y = 2$ (top) and $y = \sqrt{2} x^{1/2}$ (bottom) from $x=0$ to $x=2$.; The axis is the centroidal y-axis.; Units are in meters.