How do you set up the integral for the area moment of inertia about the y-axis using a vertical strip?

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 $x$.

Reasoning, step by step

1. Start with the general definition $I_y = \int x^2 dA$.
2. Choose a vertical differential strip of width $dx$.
3. Determine the height of the strip by subtracting the lower boundary curve from the upper boundary curve at a given $x$.
4. Express $dA$ as (height) $\times dx$.
5. Substitute the expression for $dA$ into the integral.
6. Identify the minimum and maximum $x$ values of the region to set the integration limits.

Example

For the example region bounded by $y = 2x$ and $y = \sqrt{2} x^{1/2}$ from $x=0$ to $x=2$, the height is $(2x - \sqrt{2} x^{1/2})$. The setup becomes $I_y = \int_0^2 x^2 (2x - \sqrt{2} x^{1/2}) dx$. Note: The video script shows a slightly different setup $I_y = \int_0^2 x^2 (2 - \sqrt{2} x^{1/2}) dx$ for a region bounded by $y=2$ and $y=\sqrt{2}x^{1/2}$, demonstrating the same principle of height = top - bottom.

Common misconceptions

• Forgetting to solve the boundary curve equation for $y$ if it is given implicitly (e.g., $y^2 = 2x$ must become $y = \sqrt{2} x^{1/2}$).
• Using a horizontal strip (width $dy$) when the boundaries are more easily expressed as functions of $x$.
• Incorrectly setting the integration limits based on the y-axis instead of the x-axis.

Watch the explanation

Explore next

• Why does the area moment of inertia about the y-axis use $x^2$ in the integral?
• How do you evaluate the definite integral for the area moment of inertia?
• What is the final numerical value of $I_y$ in this worked example?