Question

Find the area of the region (in square units) enclosed by the curves: \[ y^2 = 8(x+2), \quad y^2 = 4(1-x) \] and the Y-axis. 

• A. \( \frac{8}{3} (5 - 3\sqrt{2}) \)
• B. \( \frac{8}{3} (\sqrt{2} -1) \)
• C. \( \frac{8}{3} (3 - \sqrt{2}) \)
• D. \( \frac{4}{3} (\sqrt{2} + 1) \)

Hint:

When finding the area enclosed between two curves, always express them in terms of \(y^2\) or solve for \(x\), and then integrate the upper function minus the lower function over the given limits.

Answer 1

The Correct Option is A

Step 1: Identify the intersection points 
The given equations are: \[ y^2 = 8(x+2), \quad y^2 = 4(1-x). \] Equating both, \[ 8(x+2) = 4(1-x). \] Solving for \(x\), \[ 8x + 16 = 4 - 4x. \] \[ 12x = -12. \] \[ x = -1. \] For \(x = -1\), substituting in \(y^2 = 8(x+2)\), \[ y^2 = 8(-1+2) = 8. \] Thus, \(y = \pm 2\sqrt{2}\). 
Step 2: Compute the Area 
The required area is obtained using the formula: \[ A = \int_{x_1}^{x_2} (y_{\text{upper}} - y_{\text{lower}}) dx. \] After performing the necessary integration and simplifications, \[ A = \frac{8}{3} (5 - 3\sqrt{2}). \] 
Step 3: Final Answer 
Thus, the area enclosed by the given curves is: \[ \boxed{\frac{8}{3} (5 - 3\sqrt{2})}. \]