What is the final numerical value of $I_y$ in this worked example?

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.

Reasoning, step by step

1. Set up the definite integral $I_y = \int_0^2 x^2 (2 - \sqrt{2} x^{1/2}) dx$.
2. Expand the integrand to $2x^2 - \sqrt{2} x^{5/2}$.
3. Find the antiderivative: $\frac{2x^3}{3} - \sqrt{2} \frac{2}{7} x^{7/2}$.
4. Evaluate at the upper limit $x=2$: $\frac{2(2)^3}{3} - \sqrt{2} \frac{2}{7} (2)^{7/2} = \frac{16}{3} - \frac{2\sqrt{2}}{7} 2^{3.5}$.
5. Calculate the numerical values: $\frac{16}{3} \approx 5.33333$ and $\frac{2\sqrt{2}}{7} 2^{3.5} \approx 4.571$.
6. Subtract the second term from the first: $5.33333 - 4.571 = 0.762$.
7. Attach the unit $m^4$.

Example

The board shows the final calculation: $5.33 - 4.571 = .762 \text{ m}^4 = I_y$. The instructor says, "I y is equal to five point three three three three three minus answer equals point seven six two."

Common misconceptions

• Forgetting to attach the unit $m^4$ to the numerical result.
• Making arithmetic errors when evaluating the fractional exponents.
• Confusing the value of $I_y$ with the area or centroid location.

Watch the explanation

Explore next

• How do you set up the integral for the area moment of inertia about the y-axis using a vertical strip?
• Why does the area moment of inertia have units of $m^4$?
• When can you skip the parallel axis theorem in this setup?