Let's solve each question step by step. --- ## **Question 3** **Use a double integral to find the volume of the solid bounded above by the plane \( z = 4 - x - y \) and below rectangle \( R = [0,1] \times [0,3] \).** The volume \( V \) is given by the double integral: \[ V = \iint_R (4 - x - y) \, dA \] where \( R = [0,1] \times [0,3] \). Let's write this as iterated integral: \[ V = \int_{x=0}^{1} \int_{y=0}^{3} (4 - x - y) \, dy \, dx \] First, integrate with respect to \( y \): \[ \int_{y=0}^{3} (4 - x - y) \, dy = \left[ 4y - xy - \frac{y^2}{2} \right]_{y=0}^{y=3} \] Plug in \( y = 3 \) and \( y = 0 \): \[ = 4(3) - x(3) - \frac{3^2}{2} - (0 - 0 - 0) = 12 - 3x - \frac{9}{2} = 12 - 3x - 4.5 = 7.5 - 3x \] Now integrate with respect to \( x \): \[ \int_{x=0}^{1} (7.5 - 3x) \, dx = \left[ 7.5x - \frac{3x^2}{2} \right]_{x=0}^{x=1} \] Plug in \( x = 1 \) and \( x = 0 \): \[ = 7.5(1) - \frac{3(1^2)}{2} - (0 - 0) = 7.5 - 1.5 = 6 \] **Final Answer for Question 3:** \[ \boxed{6} \] --- ## **Question 4** **Evaluate the iterated integral:** \[ \int_{0}^{1} \int_{0}^{2x} e^{x^2} \, dy \, dx \] First, integrate with respect to \( y \): \( e^{x^2} \) is constant with respect to \( y \), so: \[ \int_{0}^{2x} e^{x^2} \, dy = e^{x^2} \cdot y \Big|_{y=0}^{y=2x} = e^{x^2} \cdot (2x - 0) = 2x e^{x^2} \] Now, integrate with respect to \( x \): \[ \int_{0}^{1} 2x e^{x^2} dx \] Let \( u = x^2 \) so \( du = 2x dx \). When \( x = 0, u = 0 \). When \( x = 1, u = 1 \). So the integral becomes: \[ \int_{u=0}^{u=1} e^u du = e^u \Big|_{u=0}^{u=1} = e^{1} - e^{0} = e - 1 \] Approximate \( e \approx 2.71828 \): \[ e - 1 \approx 2.72 - 1 = 1.72 \] **Final Answer for Question 4:** \[ \boxed{1.72} \] --- ### **Summary** - **Q3:** 6 - **Q4:** 1.72 Let me know if you need more explanation!