Let's solve the problem step by step. --- ### **Step 1: Sketch and Boundaries** The lamina is bounded by: - \( y = x^2 \) (above) - \( y = L \) (below, where \( L < 0 \)) - \( x = -1 \) (left) - \( x = 1 \) (right) So, the region \( R \) is: \[ R = \{ (x, y) \mid -1 \leq x \leq 1, \ L \leq y \leq x^2 \} \] --- ### **Step 2: Area of the Region (\(A\))** \[ A = \int_{x = -1}^{1} \int_{y = L}^{x^2} dy \, dx \] Integrate with respect to \( y \): \[ A = \int_{x = -1}^{1} (x^2 - L) dx \] \[ A = \int_{x = -1}^{1} x^2 dx - \int_{x = -1}^{1} L dx \] \[ \int_{-1}^{1} x^2 dx = \left[\frac{x^3}{3}\right]_{-1}^{1} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3} \] \[ \int_{-1}^{1} dx = x \Big|_{-1}^{1} = 1 - (-1) = 2 \] So, \[ A = \frac{2}{3} - 2L \] --- ### **Step 3: \( x \)-coordinate of Centre of Mass (\( \bar{x} \))** \[ \bar{x} = \frac{1}{A} \int_{x = -1}^{1} \int_{y = L}^{x^2} x \, dy \, dx \] Integrate with respect to \( y \): \[ \int_{y=L}^{x^2} x \, dy = x(y) \Big|_{L}^{x^2} = x(x^2 - L) \] So, \[ \int_{x=-1}^{1} x(x^2 - L) dx = \int_{x=-1}^{1} (x^3 - Lx) dx \] Calculate separately: - \( \int_{-1}^{1} x^3 dx = \left[ \frac{x^4}{4} \right]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0 \) - \( \int_{-1}^{1} x dx = \left[ \frac{x^2}{2} \right]_{-1}^{1} = \frac{1}{2} - \frac{1}{2} = 0 \) So, \[ \int_{x=-1}^{1} x(x^2 - L) dx = 0 - 0 = 0 \] Therefore, \[ \boxed{\bar{x} = 0} \] --- ### **Step 4: \( y \)-coordinate of Centre of Mass (\( \bar{y} \))** \[ \bar{y} = \frac{1}{A} \int_{x = -1}^{1} \int_{y = L}^{x^2} y \, dy \, dx \] Integrate with respect to \( y \): \[ \int_{y=L}^{x^2} y \, dy = \frac{1}{2} y^2 \Big|_{L}^{x^2} = \frac{1}{2} (x^4 - L^2) \] So, \[ \int_{x=-1}^{1} \frac{1}{2}(x^4 - L^2) dx = \frac{1}{2} \int_{x=-1}^{1} x^4 dx - \frac{1}{2} \int_{x=-1}^{1} L^2 dx \] Calculate: - \( \int_{-1}^{1} x^4 dx = \left[ \frac{x^5}{5} \right]_{-1}^{1} = \frac{1}{5} - (-\frac{1}{5}) = \frac{2}{5} \) - \( \int_{-1}^{1} L^2 dx = L^2 \cdot 2 \) So, \[ \int_{x=-1}^{1} \frac{1}{2}(x^4 - L^2) dx = \frac{1}{2} \times \frac{2}{5} - \frac{1}{2} \times 2 L^2 = \frac{1}{5} - L^2 \] Therefore, \[ \bar{y} = \frac{1}{A} \left( \frac{1}{5} - L^2 \right) \] Recall \( A = \frac{2}{3} - 2L \): \[ \boxed{ \bar{y} = \frac{\frac{1}{5} - L^2}{\frac{2}{3} - 2L} } \] --- ### **Step 5: For which \( L \) is the centre of mass inside the region?** The centre of mass is inside if \( L \leq \bar{y} \leq 0 \) (since \( y = L \) is the bottom, and \( y = x^2 \leq 1 \) so the max at \( (x=1) \) is 1, but the region is symmetric, so the highest point along \( x = 0 \) is \( y = 0 \)). So, we need: \[ L \leq \frac{\frac{1}{5} - L^2}{\frac{2}{3} - 2L} \leq 0 \] #### (a) When is \( \bar{y} \leq 0 \)? \[ \frac{1}{5} - L^2 \leq 0 \implies L^2 \geq \frac{1}{5} \implies L \leq -\frac{1}{\sqrt{5}} \] (since \( L < 0 \)) #### (b) When is \( \bar{y} \geq L \)? \[ \frac{\frac{1}{5} - L^2}{\frac{2}{3} - 2L} \geq L \] \[ \frac{1}{5} - L^2 \geq L \left( \frac{2}{3} - 2L \right) \] \[ \frac{1}{5} - L^2 \geq \frac{2}{3}L - 2L^2 \] \[ \frac{1}{5} - L^2 - \frac{2}{3}L + 2L^2 \geq 0 \] \[ \frac{1}{5} + L^2 - \frac{2}{3}L \geq 0 \] Multiply both sides by 15: \[ 3 + 15L^2 - 10L \geq 0 \] \[ 15L^2 - 10L + 3 \geq 0 \] Solve \( 15L^2 - 10L + 3 = 0 \): \[ L = \frac{10 \pm \sqrt{100 - 180}}{30} \] \[ = \frac{10 \pm \sqrt{-80}}{30} \] Since discriminant is negative, it's **always true** for all \( L \). But let's check for \( L < 0 \): The quadratic \( 15L^2 - 10L + 3 \) opens upwards and is always positive for all real \( L \). So, the **only restriction** is from (a): \[ L \leq -\frac{1}{\sqrt{5}} \] --- ### **Final Answers** **1. Centre of mass:** \[ \boxed{ \bar{x} = 0, \qquad \bar{y} = \frac{\frac{1}{5} - L^2}{\frac{2}{3} - 2L} } \] **2. Centre of mass is inside the region for:** \[ \boxed{ L \leq -\frac{1}{\sqrt{5}} } \] **3. Centre of mass is outside the region for:** \[ \boxed{ -\frac{1}{\sqrt{5}} < L < 0 } \] --- **If you need the centre of mass in terms of a decimal for the boundary:** \[ -\frac{1}{\sqrt{5}} \approx -0.447 \] So, for \( L \leq -0.447 \), the centre of mass is inside the region.

Let's carefully analyze and compute each part of the problem step-by-step, providing detailed explanations and calculations. --- ## **Problem Recap** - **Region:** The lamina occupies the region in the \( y \)-\( x \) plane bounded by: - \( y = x^2 \) (parabola above) - \( y = L \) (a horizontal line below, with \( L < 0 \)) - \( x = -1 \) (left boundary) - \( x = 1 \) (right boundary) - **Density:** Constant (say, \( \rho = 1 \)) - **Goal:** 1. Find the coordinates of the **center of mass** \( (\bar{x}, \bar{y}) \). 2. Determine for which values of \( L < 0 \) the center of mass lies **inside** or **outside** the region. --- ## **Step 1: Describe the Region** The region \( R \) is: \[ R = \{ (x, y) \mid -1 \leq x \leq 1, \ L \leq y \leq x^2 \} \] --- ## **Step 2: Calculate the Area \( A \)** The area is: \[ A = \iint_{R} dA = \int_{x=-1}^{1} \int_{y=L}^{x^2} dy \, dx \] Integrate with respect to \( y \): \[ A = \int_{-1}^{1} (x^2 - L) dx \] Now, compute: \[ A = \int_{-1}^1 x^2 dx - \int_{-1}^1 L dx \] \[ \int_{-1}^1 x^2 dx = \left[\frac{x^3}{3}\right]_{-1}^1 = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3} \] \[ \int_{-1}^1 dx = 2 \] Thus, \[ A = \frac{2}{3} - 2L \] --- ## **Step 3: Find \( \bar{x} \) (Center of Mass in \( x \)-direction)** \[ \bar{x} = \frac{1}{A} \iint_{R} x \, dA \] Expressed as: \[ \bar{x} = \frac{1}{A} \int_{-1}^{1} \int_{L}^{x^2} x \, dy \, dx \] Since \( x \) is independent of \( y \): \[ \int_{L}^{x^2} x \, dy = x (x^2 - L) \] Therefore, \[ \bar{x} = \frac{1}{A} \int_{-1}^{1} x (x^2 - L) dx \] Expand the integrand: \[ x(x^2 - L) = x^3 - L x \] Compute: \[ \int_{-1}^{1} x^3 dx = \left[\frac{x^4}{4}\right]_{-1}^{1} = \frac{1}{4} - \frac{1}{4} = 0 \] \[ \int_{-1}^{1} x dx = \left[\frac{x^2}{2}\right]_{-1}^{1} = \frac{1}{2} - \frac{1}{2} = 0 \] Thus, \[ \int_{-1}^{1} (x^3 - L x) dx = 0 - 0 = 0 \] Finally, \[ \boxed{ \bar{x} = 0 } \] **Interpretation:** The region is symmetric about the \( y \)-axis, so the center of mass's \( x \)-coordinate is zero. --- ## **Step 4: Find \( \bar{y} \) (Center of Mass in \( y \)-direction)** \[ \bar{y} = \frac{1}{A} \iint_{R} y \, dA \] Expressed as: \[ \bar{y} = \frac{1}{A} \int_{-1}^{1} \int_{L}^{x^2} y \, dy \, dx \] Integrate with respect to \( y \): \[ \int_{L}^{x^2} y \, dy = \frac{1}{2} y^2 \Big|_{L}^{x^2} = \frac{1}{2} (x^4 - L^2) \] Now, \[ \bar{y} = \frac{1}{A} \int_{-1}^{1} \frac{1}{2} (x^4 - L^2) dx = \frac{1}{2A} \left( \int_{-1}^{1} x^4 dx - L^2 \int_{-1}^{1} dx \right) \] Calculate the integrals: \[ \int_{-1}^{1} x^4 dx = \left[\frac{x^5}{5}\right]_{-1}^{1} = \frac{1}{5} - \left(-\frac{1}{5}\right) = \frac{2}{5} \] \[ \int_{-1}^{1} dx = 2 \] Recall \( A = \frac{2}{3} - 2L \). Putting it all together: \[ \bar{y} = \frac{1}{2A} \left( \frac{2}{5} - 2 L^2 \right) = \frac{\frac{2}{5} - 2 L^2}{2A} \] Simplify numerator: \[ \frac{2}{5} - 2 L^2 \] Thus, \[ \boxed{ \bar{y} = \frac{\frac{2}{5} - 2 L^2}{2 \left(\frac{2}{3} - 2L \right)} = \frac{\frac{2}{5} - 2 L^2}{2 \times \left(\frac{2}{3} - 2L\right)} \] Further simplify denominator: \[ 2 \times \left(\frac{2}{3} - 2L\right) = \frac{4}{3} - 4L \] **Final formula:** \[ \boxed{ \bar{y} = \frac{\frac{2}{5} - 2L^2}{\frac{4}{3} - 4L} } \] --- ## **Step 5: Determine when \( \bar{y} \) lies **inside** the region** - The region extends vertically from \( y = L \) (bottom) to \( y = x^2 \) (top). - Since the region is bounded above by \( y = x^2 \) and below by \( y = L \), the **center of mass** lies **inside** the region if: \[ L \leq \bar{y} \leq \max_{x} x^2 \] - The maximum of \( x^2 \) over \( x \in [-1, 1] \) is at \( x = \pm 1 \): \[ \max_{x} x^2 = 1 \] - The bottom boundary is at \( y = L \), which is negative, and the top boundary at \( y = x^2 \leq 1 \). **Therefore:** \[ \boxed{ L \leq \bar{y} \leq 1 } \] Since \( L < 0 \), and \( \bar{y} \) is given by the formula, examine the bounds. --- ## **Step 6: Find the critical value of \( L \) where \( \bar{y} = L \)** Set: \[ \frac{\frac{2}{5} - 2L^2}{\frac{4}{3} - 4L} = L \] Cross-multiplied: \[ \frac{2}{5} - 2L^2 = L \left( \frac{4}{3} - 4L \right) \] Simplify RHS: \[ L \times \frac{4}{3} - 4L^2 \] Bring all to one side: \[ \frac{2}{5} - 2L^2 - \frac{4}{3} L + 4L^2 = 0 \] Combine like terms: \[ \left(-2L^2 + 4L^2 \right) + \left( - \frac{4}{3} L \right) + \frac{2}{5} = 0 \] \[ (2L^2) - \frac{4}{3} L + \frac{2}{5} = 0 \] Multiply through by the least common denominator, 15: \[ 15 \times 2L^2 - 15 \times \frac{4}{3} L + 15 \times \frac{2}{5} = 0 \] Calculate: \[ 30 L^2 - 20 L + 6 = 0 \] Divide through by 2: \[ 15 L^2 - 10 L + 3 = 0 \] Solve for \( L \): \[ L = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 15 \times 3}}{2 \times 15} = \frac{10 \pm \sqrt{100 - 180}}{30} \] \[ L = \frac{10 \pm \sqrt{-80}}{30} \] Since discriminant is negative, **no real solutions**. **Implication:** - \( \bar{y} \) **never equals** \( L \) for real \( L \), meaning the center of mass always **lies above** the bottom boundary \( y=L \). --- ## **Step 7: Check when \( \bar{y} \leq 1 \)** For the center of mass to be **inside** the region, we require: \[ \bar{y} \leq 1 \] Compute: \[ \frac{\frac{2}{5} - 2L^2}{\frac{4}{3} - 4L} \leq 1 \] Cross-multiplied (denominator positive?): - For \( L < 0 \), numerator is generally positive for small \( |L| \), but to be safe, analyze the sign. Assuming the denominator \( \frac{4}{3} - 4L \) > 0: \[ \frac{4}{3} - 4L > 0 \implies L < \frac{1}{3} \] which is true since \( L < 0 \). So, multiplying both sides by positive denominator: \[ \frac{2}{5} - 2L^2 \leq \frac{4}{3} - 4L \] Bring all to LHS: \[ \frac{2}{5} - 2L^2 - \frac{4}{3} + 4L \leq 0 \] Combine constants: \[ \left(\frac{2}{5} - \frac{4}{3}\right) + 4L - 2L^2 \leq 0 \] Calculate: \[ \frac{2}{5} - \frac{4}{3} = \frac{6}{15} - \frac{20}{15} = -\frac{14}{15} \] So inequality: \[ -\frac{14}{15} + 4L - 2L^2 \leq 0 \] Multiply through by 15: \[ -14 + 60 L - 30 L^2 \leq 0 \] Rearranged: \[ -30 L^2 + 60 L - 14 \leq 0 \] Multiply through by -1 (reverse inequality): \[ 30 L^2 - 60 L + 14 \geq 0 \] Divide through by 2: \[ 15 L^2 - 30 L + 7 \geq 0 \] --- ## **Step 8: Solve quadratic inequality** Quadratic: \[ 15 L^2 - 30 L + 7 \geq 0 \] Discriminant: \[ \Delta = (-30)^2 - 4 \times 15 \times 7 = 900 - 420 = 480 \] Square root: \[ \sqrt{480} = \sqrt{16 \times 30} = 4 \sqrt{30} \approx 4 \times 5.477 = 21.908 \] Solutions: \[ L = \frac{30 \pm \sqrt{480}}{2 \times 15} = \frac{30 \pm 4 \sqrt{30}}{30} = 1 \pm \frac{2 \sqrt{30}}{15} \] Numerically: \[ \frac{2 \sqrt{30}}{15} \approx \frac{2 \times 5.477}{15} \approx \frac{10.954}{15} \approx 0.730 \] Thus, \[ L = 1 \pm 0.730 \] - \( L_1 = 1 - 0.730 \approx 0.27 \

Let's continue from where we left off. --- ## **Step 8 (continued): Numerical solutions of the quadratic inequality** Quadratic: \[ 15 L^2 - 30 L + 7 \geq 0 \] Solutions: \[ L = \frac{30 \pm \sqrt{480}}{30} \] Calculate \( \sqrt{480} \): \[ \sqrt{480} = \sqrt{16 \times 30} = 4 \sqrt{30} \approx 4 \times 5.477 = 21.908 \] Therefore, \[ L = \frac{30 \pm 21.908}{30} \] Compute the two roots: 1. **First root:** \[ L_{1} = \frac{30 + 21.908}{30} = \frac{51.908}{30} \approx 1.730 \] 2. **Second root:** \[ L_{2} = \frac{30 - 21.908}{30} = \frac{8.092}{30} \approx 0.2707 \] --- ## **Step 9: Interpretation of the quadratic inequality** Since the quadratic coefficient (15) is positive, the parabola opens upward. The inequality: \[ 15 L^2 - 30 L + 7 \geq 0 \] holds for: \[ L \leq L_{2} \quad \text{or} \quad L \geq L_{1} \] But note the roots: - \( L_{2} \approx 0.2707 \) - \( L_{1} \approx 1.730 \) Given that \( L < 0 \) (by problem statement), the relevant part is: \[ L \leq L_2 \approx 0.2707 \] Since \( L < 0 \), all negative \( L \) satisfy: \[ L \leq 0.2707 \] which is always true for \( L < 0 \). **Conclusion:** - The condition \( \bar{y} \leq 1 \) is always satisfied for all negative \( L \). --- ## **Summary of the Conditions** - The center of mass \( (\bar{x}, \bar{y}) \) always has: \[ \bar{x} = 0 \] \[ \bar{y} = \frac{\frac{2}{5} - 2 L^2}{\frac{4}{3} - 4L} \] - The **region bounds**: - \( y \) is between \( L \) (bottom) and \( x^2 \) (top). - The maximum of \( x^2 \) over \( x \in [-1, 1] \) is 1. --- ## **Final answer:** ### **Center of mass:** \[ \boxed{ \boxed{ \text{Coordinates:} \quad (\bar{x}, \bar{y}) = \left( 0, \frac{\frac{2}{5} - 2L^2}{\frac{4}{3} - 4L} \right) } } \] --- ### **Range of \( L \) for the center of mass to be inside the region:** - Since \( y \) of the center of mass is always greater than or equal to \( L \) (as shown by the inequality analysis), and less than or equal to 1 (region's maximum height), the center of mass is **inside the region** if: \[ L \leq \bar{y} \leq 1 \] - Because \( \bar{y} \) is always greater than \( L \) for \( L < 0 \), and less than 1, **the center of mass lies inside the region for all \( L < 0 \) such that** \[ L \leq \text{some critical value} \] - From earlier calculations, the critical value where \( \bar{y} = L \) **does not** have real solutions, meaning: \[ \boxed{ \text{The center of mass always lies inside the region for all } L < 0 } \] --- ## **Summary:** | **Parameter** | **Condition** | **Implication** | | --- | --- | --- | | \( \bar{x} = 0 \) | Always | Due to symmetry | | \( \bar{y} = \frac{\frac{2}{5} - 2L^2}{\frac{4}{3} - 4L} \) | Always defined for \( L < 0 \) | Center of mass is located at this \( y \)-coordinate | | Inside region? | \( L \leq \bar{y} \leq 1 \) | Always true for \( L < 0 \) (since \( \bar{y} > L \) and \( \bar{y} < 1 \)) | **Thus, the center of mass always lies inside the region for all \( L < 0 \).** --- **Note:** If more precise bounds are desired, numerical evaluation can be used, but the overall conclusion remains that the center of mass is within the region for all \( L < 0 \). --- **Feel free to ask for further clarifications!**