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

Given the equations \( x(1 + y^2) = 1 \) and \( y^2 = 2x \), we find the area of the region bounded by these curves. First, we solve for the points of intersection.
Step 1: Solve the system of equations. From \( y^2 = 2x \), we get \( x = \frac{y^2}{2} \). Substituting this into the first equation gives \( \frac{y^2}{2}(1 + y^2) = 1 \), which simplifies to \( y^4 + y^2 - 2 = 0 \). Let \( z = y^2 \). The equation becomes \( z^2 + z - 2 = 0 \). Using the quadratic formula, \( z = \frac{-1 \pm \sqrt{1 - 4(1)(-2)}}{2} = \frac{-1 \pm 3}{2} \). Thus, \( z = 1 \) or \( z = -2 \). We reject \( z = -2 \) since \( y^2 \geq 0 \). Therefore, \( y^2 = 1 \), which means \( y = \pm 1 \).
Step 2: Calculate the area. The area is calculated by integrating the difference between the two curves with respect to \( y \): \[ A = \int_{-1}^{1} \left( x_2 - x_1 \right) \, dy \] where \( x_2 = \frac{1}{1 + y^2} \) and \( x_1 = \frac{y^2}{2} \). Evaluating the integral yields: \[ A = \frac{\pi}{2} - \frac{1}{3} \]