Question

The area of the region bounded by the curves \( x(1 + y^2) = 1 \) and \( y^2 = 2x \) is:

• A. \( \frac{\pi}{4} - \frac{1}{3} \)
• B. \( \frac{\pi}{2} - \frac{1}{3} \)
• C. \( \frac{1}{2}[\frac{\pi}{2} - \frac{1}{3}] \)
• D. \( 2[\frac{\pi}{2} - \frac{1}{3}] \)

Hint:

When solving for the area between curves, carefully set up the integral and use the limits of intersection.

Answer 1

The Correct Option is B

We are given the equations \( x(1 + y^2) = 1 \) and \( y^2 = 2x \). To find the area of the region bounded by these curves, we first solve for the points of intersection. 
Step 1: Solve the system of equations. From the second equation, solve for \( x \): \[ y^2 = 2x \quad \Rightarrow \quad x = \frac{y^2}{2} \] Substitute this into the first equation: \[ \frac{y^2}{2}(1 + y^2) = 1 \quad \Rightarrow \quad y^2 + y^4 = 2 \] This simplifies to: \[ y^4 + y^2 - 2 = 0 \] Let \( z = y^2 \), so we have: \[ z^2 + z - 2 = 0 \] Solve for \( z \) using the quadratic formula: \[ z = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \] Thus, \( z = 1 \) or \( z = -2 \) (reject \( z = -2 \) because \( y^2 \geq 0 \)). So, \( y^2 = 1 \), hence \( y = \pm 1 \). 
Step 2: Calculate the area. The area is given by the integral of the difference between the two curves: \[ A = \int_{-1}^{1} \left( x_2 - x_1 \right) \, dy \] where \( x_2 = \frac{y^2}{2} \) and \( x_1 = \frac{1}{1 + y^2} \). Calculate the integral and find: \[ A = \frac{\pi}{2} - \frac{1}{3} \]