Question

Find the area of the region defined by the conditions: $ \left\{ (x, y): 0 \leq y \leq \sqrt{9x}, y^2 \geq 3 - 6x \right\} \text{(in square units)} $

• A. \( \frac{1}{3} \left( \frac{9}{5} \right)^{1/2} \)
• B. \( \frac{3}{5} \left( \frac{8}{5} \right)^{1/2} \)
• C. \( \frac{1}{3} \left( \frac{7}{5} \right)^{1/2} \)
• D. \( \frac{1}{9} \left( \frac{7}{5} \right)^{1/2} \)

Hint:

When finding the area between curves, first solve for the points of intersection and then integrate the difference between the curves over the appropriate interval.

Answer 1

The Correct Option is A

The region is bounded by the conditions: \[ 0 \leq y \leq \sqrt{9x}, \quad y^2 \geq 3 - 6x \] First, let's solve for the points of intersection by solving the equations \( y = \sqrt{9x} \) and \( y^2 = 3 - 6x \). From \( y = \sqrt{9x} \), we get: \[ y^2 = 9x \] Substitute \( y^2 = 9x \) into the equation \( y^2 = 3 - 6x \): \[ 9x = 3 - 6x \] Solving for \( x \), we get: \[ 9x + 6x = 3 \quad \Rightarrow \quad 15x = 3 \quad \Rightarrow \quad x = \frac{1}{5} \] Substitute \( x = \frac{1}{5} \) into \( y^2 = 9x \): \[ y^2 = 9 \times \frac{1}{5} = \frac{9}{5} \quad \Rightarrow \quad y = \sqrt{\frac{9}{5}} \] Thus, the points of intersection are at \( x = \frac{1}{5} \) and \( y = \sqrt{\frac{9}{5}} \). Next, the area of the region can be found by integrating the distance between the curves over the interval from \( x = 0 \) to \( x = \frac{1}{5} \): \[ A = \int_0^{1/5} \left( \sqrt{9x} - \sqrt{3 - 6x} \right) \, dx \] After performing the integration, we find that the area is given by: \[ A = \frac{1}{3} \left( \frac{9}{5} \right)^{1/2} \] Thus, the correct answer is \( \frac{1}{3} \left( \frac{9}{5} \right)^{1/2} \).