Certainly! To solve these problems using direct integration, we need to understand the shape and set up the integrals accordingly. Since the shape is not explicitly provided here, I will assume a common shape for demonstration, such as a rectangle, and specify the process generally. If you have a specific shape, please provide its dimensions or a diagram. --- ### Assumptions: - Shape: Rectangular area with width \(b\) and height \(h\). - Point \(P\): Located at a reference point within or on the shape; for simplicity, assume \(P\) is at the centroid or at a specific coordinate (e.g., origin). --- ## Part (a): Moment of inertia about horizontal x-axis through point \(P\) ### Step 1: Coordinate system setup - Place the shape such that its centroid is at the origin \(O(0,0)\). - The x-axis is horizontal, y-axis is vertical. - The differential element \(dA\) is a horizontal strip at coordinate \(y\). ### Step 2: Differential area element - For a rectangle: \(dA = b\,dy\), where \(y\) varies from \(-h/2\) to \(h/2\). - If \(P\) is at the centroid, the moment of inertia about the x-axis (about the centroid) is: \[ I_x = \int_A y^2\, dA \] ### Step 3: Set up the integral \[ I_x = \int_{y=-h/2}^{h/2} y^2 \cdot b\, dy \] ### Step 4: Integrate \[ I_x = b \int_{-h/2}^{h/2} y^2\, dy \] \[ I_x = b \left[ \frac{y^3}{3} \right]_{-h/2}^{h/2} \] \[ I_x = b \left( \frac{(h/2)^3 - (-h/2)^3}{3} \right) \] Note that \(y^3\) is odd, so: \[ I_x = b \cdot \frac{2 \times (h/2)^3}{3} = b \cdot \frac{2 \times h^3/8}{3} = b \cdot \frac{h^3/4}{3} = \frac{b h^3}{12} \] ### **Answer for (a):** \[ \boxed{ I_x = \frac{b h^3}{12} } \] --- ## Part (b): Moment of inertia about vertical y-axis through point \(P\) ### Step 1: Set up similar to part (a) - For a rectangle, the differential element is a vertical strip at \(x\), with width \(dx\). - \(dA = h\, dx\), with \(x\) from \(-b/2\) to \(b/2\). ### Step 2: Set up the integral \[ I_y = \int_A x^2\, dA \] \[ I_y = \int_{x=-b/2}^{b/2} x^2 \cdot h\, dx \] ### Step 3: Integrate \[ I_y = h \int_{-b/2}^{b/2} x^2\, dx \] \[ I_y = h \left[ \frac{x^3}{3} \right]_{-b/2}^{b/2} \] \[ I_y = h \left( \frac{(b/2)^3 - (-b/2)^3}{3} \right) \] Since \(x^3\) is odd: \[ I_y = h \cdot \frac{2 \times (b/2)^3}{3} = h \cdot \frac{2 \times b^3/8}{3} = h \cdot \frac{b^3/4}{3} = \frac{h b^3}{12} \] ### **Answer for (b):** \[ \boxed{ I_y = \frac{h b^3}{12} } \] --- ## Part (c): Polar Moment of Inertia about point \(P\) and radius of gyration ### Step 1: Polar moment of inertia \(J_P\) For a rectangular section centered at the origin: \[ J_P = I_x + I_y = \frac{b h^3}{12} + \frac{h b^3}{12} = \frac{b h^3 + h b^3}{12} \] ### Step 2: Radius of gyration \(r_g\) \[ r_g = \sqrt{\frac{J_P}{A}} \] where \(A = b \times h\). ### Final expressions: \[ \boxed{ J_P = \frac{b h^3 + h b^3}{12} } \] \[ \boxed{ r_g = \sqrt{\frac{J_P}{A}} = \sqrt{\frac{\frac{b h^3 + h b^3}{12}}{b h}} = \sqrt{\frac{b h^3 + h b^3}{12 b h}} \] Simplify numerator: \[ r_g = \sqrt{\frac{b h^3 + h b^3}{12 b h}} = \sqrt{\frac{b h^3}{12 b h} + \frac{h b^3}{12 b h}} = \sqrt{\frac{h^2}{12} + \frac{b^2}{12}} = \frac{1}{2 \sqrt{3}} \sqrt{b^2 + h^2} \] --- ## **Summary:** | Quantity | Expression | |---|---| | (a) \(I_x\) | \(\displaystyle \frac{b h^3}{12}\) | | (b) \(I_y\) | \(\displaystyle \frac{h b^3}{12}\) | | (c) \(J_P\) | \(\displaystyle \frac{b h^3 + h b^3}{12}\) | | Radius of gyration \(r_g\) | \(\displaystyle \frac{\sqrt{b^2 + h^2}}{2 \sqrt{3}}\) | --- **Note:** If your shape differs from a rectangle or if point \(P\) is located elsewhere, please specify the shape and location for a tailored solution.