## Problem Restatement **Given:** - Thin rectangular plate on the xy-plane - Side lengths: \( a = .3 \) m, \( b = .4 \) m - Surface mass density: \( \sigma(x, y) = 8xy \) kg/m² - Find the **moment of inertia about the z-axis** (perpendicular to the plate through the origin). --- ## **Step 1: Set Up the Integral for Moment of Inertia** The **moment of inertia** about the z-axis is: \[ I_z = \iint_{A} r^2 \, dm \] Where \( r^2 = x^2 + y^2 \), and \( dm = \sigma(x, y) \, dx\,dy \). So, \[ I_z = \int_^a \int_^b (x^2 + y^2) \cdot 8xy \, dy\,dx \] --- ## **Step 2: Calculate the Moment of Inertia** Let's expand and compute the double integral: \[ I_z = 8 \int_^a \int_^b (x^2 + y^2)xy \, dy\,dx \] \[ = 8 \int_^a \int_^b (x^3y + x y^3) \, dy\,dx \] ### **Integrate with respect to \( y \):** \[ \int_^b (x^3 y + x y^3) \, dy = x^3 \int_^b y \, dy + x \int_^b y^3 \, dy \] \[ = x^3 \left[ \frac{y^2}{2} \right]_^b + x \left[ \frac{y^4}{4} \right]_^b \] \[ = x^3 \frac{b^2}{2} + x \frac{b^4}{4} \] ### **Integrate with respect to \( x \):** \[ I_z = 8 \int_^a \left( x^3 \frac{b^2}{2} + x \frac{b^4}{4} \right) dx \] \[ = 8 \left( \frac{b^2}{2} \int_^a x^3 dx + \frac{b^4}{4} \int_^a x dx \right) \] Calculate the integrals: - \( \int_^a x^3 dx = \frac{a^4}{4} \) - \( \int_^a x dx = \frac{a^2}{2} \) Plug in: \[ I_z = 8 \left( \frac{b^2}{2} \cdot \frac{a^4}{4} + \frac{b^4}{4} \cdot \frac{a^2}{2} \right) \] \[ = 8 \left( \frac{b^2 a^4}{8} + \frac{b^4 a^2}{8} \right) \] \[ = 8 \cdot \frac{b^2 a^4 + b^4 a^2}{8} \] \[ = b^2 a^4 + b^4 a^2 \] Plug in \( a = .3 \), \( b = .4 \): - \( b^2 a^4 = (.4)^2 (.3)^4 = .16 \times .0081 = .001296 \) - \( b^4 a^2 = (.4)^4 (.3)^2 = .0256 \times .09 = .002304 \) \[ I_z = .001296 + .002304 = .0036 \text{ kg m}^2 \] --- ### **Final Answer** \[ \boxed{.0036 \text{ kg m}^2} \] **So, the answer is D) .0036** --- ### **Graphical Illustration**  *Alt text: Diagram of a rectangular plate in the xy-plane with sides a and b, showing the origin and z-axis perpendicular to the plate.* --- ### **Summary Table** | Step | Description | Calculation | Value | |------|-------------|-------------|-------| | 1 | Set up double integral | \(I_z = \iint (x^2 + y^2) \sigma(x, y) dA\) | | | 2 | Solve | \(I_z = .0036 \text{ kg m}^2\) | | --- **Example:** If \( a = .5 \) m, \( b = .6 \) m, \[ I_z = (.6^2 \times .5^4) + (.6^4 \times .5^2) = (.36 \times .0625) + (.1296 \times .25) = .0225 + .0324 = .0549 \text{ kg m}^2 \]