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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 $x$.

To use the curve equation $y^2 = 2x$ in an integral setup with vertical strips (where the height must be a function of $x$), you must solve for $y$. Taking the square root of both sides gives $y = \pm\sqrt{2x}$.

Conditions: The boundary curve is given as $y^2 = 2x$.; The region is in the first quadrant ($x \ge 0, y \ge 0$).; Vertical strips are being used, requiring height as a function of $x$.

To use the curve equation $y^2 = 2x$ in an integral setup with vertical strips (where the height must be a function of $x$), you must solve for $y$. Taking the square root of both sides gives $y = \pm\sqrt{2x}$.

Conditions: The boundary curve is given as $y^2 = 2x$.; The region is in the first quadrant ($x \ge 0, y \ge 0$).; Vertical strips are being used, requiring height as a function of $x$.