Let's answer **Question 10** in two steps, with explanations: --- ### Step 1: ** if \( f(x, y) \geq \) everywhere** - The function \( f(x, y) = 4xy \) for \( (x, y) \in [,1] \times [,1] \) and \( f(x, y) = \) otherwise. - Since for \( x, y \in [,1]), both \( x \geq \) and \( y \geq \), so \( 4xy \geq \). - **Explanation:** The function is non-negative everywhere, satisfying the first property of a joint probability density function. --- ### Step 2: **Check if the integral over the entire domain equals 1** - Compute the double integral over the domain: \[ \int_^1 \int_^1 4xy \, dx\,dy \] - Integrate with respect to \( x \) first: \[ \int_^1 \left[ 4y \int_^1 x \, dx \right] dy = \int_^1 4y \left( \frac{1}{2} \right) dy = \int_^1 2y\,dy = \left[ y^2 \right]_^1 = 1 \] - **Explanation:** The total integral is 1, so the function is properly normalized. --- ### **Final Answer** **Yes, \( f(x, y) \) is a joint probability density function because it is non-negative everywhere and integrates to 1 over its domain.**

# Solutions to the Given Questions ## Question 7: Mass of a Lamina ### Step 1: Set up the integral The mass \( M \) of the lamina can be calculated using the formula: \[ M = \int \int_D p(x,y) \, dA \] Given the density function \( p(x,y) = ky \) and the region bounded by \( y = 1 - 2x \) and \( y = 0 \), we need to find the limits of integration for \( x \) and \( y \). ### Explanation The line \( y = 1 - 2x \) intersects the x-axis at \( x = 0.5 \). Thus, the limits are: - \( y \) ranges from \( 0 \) to \( 1 - 2x \) - \( x \) ranges from \( 0 \) to \( 0.5 \) So, the integral becomes: \[ M = \int_0^{0.5} \int_0^{1 - 2x} ky \, dy \, dx \] --- ## Question 8: Moment of Inertia \( I \) ### Step 1: Set up the integral The moment of inertia \( I \) for a lamina with density proportional to the square of the distance from the vertex can be expressed as: \[ I = \int \int_D r^2 \, p(x,y) \, dA \] where \( r \) is the distance from the vertex and \( p(x,y) = k(x^2 + y^2) \). ### Explanation For an isosceles right triangle, the limits for \( x \) and \( y \) are: - \( x \) ranges from \( 0 \) to \( a \) - \( y \) ranges from \( 0 \) to \( a - x \) Thus, the integral becomes: \[ I = \int_0^a \int_0^{a-x} k(x^2 + y^2) \, dy \, dx \] --- ## Question 9: Surface Area Calculation ### Step 1: Set up the integral The surface area \( S \) of the surface \( z = 2x + y \) in the first octant below \( z = 5 \) can be calculated using the formula: \[ S = \iint_D \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dA \] Here, \( \frac{\partial z}{\partial x} = 2 \) and \( \frac{\partial z}{\partial y} = 1 \). ### Explanation Thus, we have: \[ S = \iint_D \sqrt{1 + 2^2 + 1^2} \, dA = \iint_D \sqrt{6} \, dA \] The region \( D \) is bounded by \( x \geq 0 \), \( y \geq 0 \), and \( z = 5 \) leading to: \[ S = \int_0^{5/2} \int_0^{5 - 2x} \sqrt{6} \, dy \, dx \] --- ## Question 10: Joint Probability Density Function ### Step 1: Check non-negativity The function given is: \[ f(x, y) = \begin{cases} 4xy & \text{for } (x, y) \in [0, 1] \times [0, 1] \\ 0 & \text{otherwise} \end{cases} \] We check if \( f(x, y) \geq 0 \). ### Explanation Since \( x \) and \( y \) are both in the interval \([0, 1]\), it follows that \( 4xy \geq 0 \). ### Step 2: Check normalization Now, calculate: \[ \int_0^1 \int_0^1 4xy \, dx \, dy \] ### Explanation Calculating the integral, we find: \[ \int_0^1 \left( 4y \int_0^1 x \, dx \right) dy = \int_0^1 2y \, dy = 1 \] ### **Final Answer** **Yes, \( f(x, y) \) is a joint probability density function because it is non-negative everywhere and integrates to 1 over its domain.**