Question

The area of the region \( \{(x, y): |x - y| \le y \le 4\sqrt{x}\} \) is

• A. 512
• B. \( \frac{1024}{3} \)
• C. \( \frac{512}{3} \)
• D. \( \frac{2048}{3} \)

Hint:

To find the area of a region defined by inequalities, first determine the bounding curves and their intersection points. Then, set up the definite integral of the difference between the upper and lower functions over the interval defined by the intersection points. Remember to handle absolute value inequalities by splitting them into separate cases.

Answer 1

The Correct Option is B

\begin{center} \includegraphics{11S.png} \end{center} The region is defined by the inequalities: \[ |x - y| \le y \quad \text{and} \quad y \le 4\sqrt{x} \] The first inequality \( |x - y| \le y \) can be written as: \[ -y \le x - y \le y \] This splits into two inequalities: \[ -y \le x - y \implies 0 \le x \] \[ x - y \le y \implies x \le 2y \implies y \ge \frac{x}{2} \] So, the region is bounded by \( y \ge \frac{x}{2} \), \( y \le 4\sqrt{x} \), and \( x \ge 0 \). To find the intersection points of the curves \( y = \frac{x}{2} \) and \( y = 4\sqrt{x} \), we set them equal: \[ \frac{x}{2} = 4\sqrt{x} \] \[ x = 8\sqrt{x} \] Squaring both sides: \[ x^2 = 64x \] \[ x^2 - 64x = 0 \] \[ x(x - 64) = 0 \] The solutions are \( x = 0 \) and \( x = 64 \). The corresponding \( y \) values are \( y = 0 \) and \( y = \frac{64}{2} = 32 \). The intersection points are \( (0, 0) \) and \( (64, 32) \). The area of the region can be found by integrating the difference between the upper and lower bounds of \( y \) with respect to \( x \) from \( 0 \) to \( 64 \): \[ \text{Area} = \int_{0}^{64} \left( 4\sqrt{x} - \frac{x}{2} \right) dx \] \[ \text{Area} = \int_{0}^{64} \left( 4x^{1/2} - \frac{1}{2}x \right) dx \] \[ \text{Area} = \left[ 4 \frac{x^{3/2}}{3/2} - \frac{1}{2} \frac{x^2}{2} \right]_{0}^{64} \] \[ \text{Area} = \left[ \frac{8}{3} x^{3/2} - \frac{1}{4} x^2 \right]_{0}^{64} \] \[ \text{Area} = \left( \frac{8}{3} (64)^{3/2} - \frac{1}{4} (64)^2 \right) - \left( \frac{8}{3} (0)^{3/2} - \frac{1}{4} (0)^2 \right) \] \[ \text{Area} = \frac{8}{3} (8^2)^{3/2} - \frac{1}{4} (4096) \] \[ \text{Area} = \frac{8}{3} (8^3) - 1024 \] \[ \text{Area} = \frac{8}{3} (512) - 1024 \] \[ \text{Area} = \frac{4096}{3} - \frac{3072}{3} \] \[ \text{Area} = \frac{1024}{3} \]