Certainly! Let's solve the problem step-by-step. --- # Given Data: - \( y^2 = \frac{9}{2} x^2 \) - \( x = 2 \) --- # Step 1: Understand the relationship The equation relates \( y^2 \) to \( x^2 \): \[ y^2 = \frac{9}{2} x^2 \] --- # Step 2: Find the values of \( y \) at \( x=2 \) Substitute \( x=2 \) into the equation: \[ y^2 = \frac{9}{2} \times (2)^2 \] \[ y^2 = \frac{9}{2} \times 4 \] \[ y^2 = \frac{9 \times 4}{2} = \frac{36}{2} = 18 \] \[ \Rightarrow y = \pm \sqrt{18} = \pm 3\sqrt{2} \] --- # Step 3: Find the second moments of area \( I_x \) and \( I_y \) ### For \( I_x \): The second moment of area about the x-axis is given by: \[ I_x = \int y^2 \, dA \] Since the region is bounded by the curve, the second moment can be calculated over the area enclosed. ### For \( I_y \): Similarly, the second moment of area about the y-axis is: \[ I_y = \int x^2 \, dA \] --- # Step 4: Recognize the symmetry and limits The boundary is defined by: \[ y^2 = \frac{9}{2} x^2 \] which simplifies to: \[ |y| = \sqrt{\frac{9}{2}} x = \frac{3}{\sqrt{2}} x \] The region is symmetric about both axes, so we can consider positive parts and multiply accordingly. --- # Step 5: Calculate \( I_x \) and \( I_y \) ### For \( I_x \): Because \( y^2 = \frac{9}{2} x^2 \), at each \( x \), \[ y = \pm \frac{3}{\sqrt{2}} x \] and the width of the region in y is from \(-\frac{3}{\sqrt{2}} x\) to \(+\frac{3}{\sqrt{2}} x\). The differential area element: \[ dA = 2 y \, dx = 2 \times \frac{3}{\sqrt{2}} x \, dx = \frac{6}{\sqrt{2}} x \, dx \] but more simply, since the region is symmetric in y, integrating with respect to x from \( x= \) to some boundary. However, for the second moment about x-axis, the integrand is \( y^2 \, dA \): \[ I_x = \int y^2 \, dA \] Given the symmetry, the total \( I_x \) over the entire region is twice the integral from to the boundary. --- ### Final step: Since the problem asks for the second moments of area **around the axes** of the **region** bounded by \( y^2 = \frac{9}{2} x^2 \), and the boundary is linear, the moments can be computed directly using standard formulas for a region bounded between two lines or for a triangle-like region. --- # **Final Answer:** **For the second moment of area:** - \( I_x \) is proportional to the integral of \( y^2 \) over the region. - \( I_y \) is proportional to the integral of \( x^2 \) over the region. Given the symmetry and boundary, the detailed integral calculations involve setting integration limits based on the boundary line and integrating accordingly. --- **Note:** Without specific bounds or more detailed region description, the general approach is to express the region mathematically, set up the integrals for \( I_x \) and \( I_y \), evaluate them, and then substitute the bounds. --- **Summary:** - For the given boundary \( y^2 = \frac{9}{2} x^2 \), at \( x=2 \), \( y = \pm 3\sqrt{2} \). - The moments of inertia \( I_x \) and \( I_y \) are computed via integrals over the region. - Symmetry simplifies calculations, focusing on one quadrant and doubling results. --- If you'd like, I can help guide through the detailed integral setup and calculation!