Let's solve the problem step by step: ### **Step 1: Set up the equations** We have two curves: - \( x = 2y \) - \( x = y^2 - 3 \) ### **Step 2: Find intersection points** Set the equations equal to each other to find the y-values where they intersect: \[ 2y = y^2 - 3 \] \[ y^2 - 2y - 3 = 0 \] This is a quadratic equation. Factor: \[ (y - 3)(y + 1) = 0 \] So, \[ y = 3 \quad \text{and} \quad y = -1 \] ### **Step 3: Set up the area integral** The area between the curves from \( y = -1 \) to \( y = 3 \) is: \[ \text{Area} = \int_{-1}^{3} [\text{Right Curve} - \text{Left Curve}]\, dy \] Here, the right curve is \( x = 2y \), and the left curve is \( x = y^2 - 3 \): \[ \text{Area} = \int_{-1}^{3} [2y - (y^2 - 3)]\, dy \] \[ = \int_{-1}^{3} [2y - y^2 + 3]\, dy \] ### **Step 4: Integrate** Integrate term by term: \[ \int_{-1}^{3} (2y - y^2 + 3)\, dy = \left[ y^2 - \frac{y^3}{3} + 3y \right]_{-1}^{3} \] ### **Step 5: Evaluate at bounds** First, at \( y = 3 \): \[ (3)^2 - \frac{(3)^3}{3} + 3(3) = 9 - \frac{27}{3} + 9 = 9 - 9 + 9 = 9 \] Now, at \( y = -1 \): \[ (-1)^2 - \frac{(-1)^3}{3} + 3(-1) = 1 - \left(-\frac{1}{3}\right) - 3 = 1 + \frac{1}{3} - 3 = -2 + \frac{1}{3} = -\frac{5}{3} \] ### **Step 6: Subtract lower from upper** \[ \text{Area} = 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{32}{3} \] --- ## **Final Answer** \[ \boxed{\frac{32}{3}} \] So, the area of the shaded region is **\(\frac{32}{3}\)**.