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

The problem asks for the area of the region defined by the set of points \((x, y)\) that satisfy the inequalities \(|x - y| \le y \le 4\sqrt{x}\).

Concept Used:

To find the area of the region, we first need to determine the boundaries defined by the given inequalities. The area between two curves \(y = f(x)\) (upper curve) and \(y = g(x)\) (lower curve) from \(x = a\) to \(x = b\) is calculated using the definite integral:

\[ A = \int_{a}^{b} [f(x) - g(x)] \,dx \]

We will need to solve the inequalities to identify \(f(x)\), \(g(x)\), and the integration limits \(a\) and \(b\).

Step-by-Step Solution:

Step 1: Analyze the inequalities to define the boundaries of the region.

The given condition is \(|x - y| \le y \le 4\sqrt{x}\). This can be broken down into two separate inequalities:

1. \(|x - y| \le y\)
2. \(y \le 4\sqrt{x}\)

Let's analyze the first inequality, \(|x - y| \le y\). By the definition of absolute value, this is equivalent to:

\[ -y \le x - y \le y \]

This gives us two conditions:

a) \(-y \le x - y \implies 0 \le x\). So, \(x\) must be non-negative.

b) \(x - y \le y \implies x \le 2y \implies y \ge \frac{x}{2}\).

So, the inequality \(|x - y| \le y\) simplifies to the region where \(y \ge \frac{x}{2}\) and \(x \ge 0\).

The second inequality is \(y \le 4\sqrt{x}\). For \(4\sqrt{x}\) to be a real number, we must have \(x \ge 0\), which is consistent with our finding from the first inequality.

Combining all conditions, the region is bounded by the line \(y = \frac{x}{2}\) from below and the curve \(y = 4\sqrt{x}\) from above, for \(x \ge 0\).

Step 2: Find the points of intersection of the boundary curves.

To find the limits of integration, we set the expressions for the lower and upper boundaries equal to each other:

\[ \frac{x}{2} = 4\sqrt{x} \]

One obvious intersection point is at \(x=0\), which gives \(y=0\). For other points, assuming \(x > 0\), we can square both sides:

\[ \left(\frac{x}{2}\right)^2 = (4\sqrt{x})^2 \] \[ \frac{x^2}{4} = 16x \] \[ x^2 = 64x \] \[ x^2 - 64x = 0 \implies x(x - 64) = 0 \]

This gives two solutions for \(x\): \(x = 0\) and \(x = 64\).

The corresponding y-values are:

• If \(x=0\), \(y=0\). Point is (0, 0).
• If \(x=64\), \(y = \frac{64}{2} = 32\). Point is (64, 32).

The limits of integration for \(x\) are from 0 to 64.

Step 3: Set up the definite integral for the area.

The area (A) is the integral of the upper curve minus the lower curve, from \(x=0\) to \(x=64\).

\[ A = \int_{0}^{64} \left( 4\sqrt{x} - \frac{x}{2} \right) \,dx \]

Step 4: Evaluate the integral.

\[ A = \left[ 4 \frac{x^{3/2}}{3/2} - \frac{1}{2} \frac{x^2}{2} \right]_{0}^{64} \] \[ A = \left[ \frac{8}{3} x^{3/2} - \frac{x^2}{4} \right]_{0}^{64} \]

Now, substitute the limits of integration:

\[ A = \left( \frac{8}{3} (64)^{3/2} - \frac{(64)^2}{4} \right) - \left( \frac{8}{3} (0)^{3/2} - \frac{(0)^2}{4} \right) \] \[ A = \left( \frac{8}{3} (\sqrt{64})^3 - \frac{4096}{4} \right) - 0 \] \[ A = \frac{8}{3} (8)^3 - 1024 \] \[ A = \frac{8}{3} (512) - 1024 \] \[ A = \frac{4096}{3} - 1024 \]

To subtract, find a common denominator:

\[ A = \frac{4096}{3} - \frac{3 \times 1024}{3} = \frac{4096 - 3072}{3} \] \[ A = \frac{1024}{3} \]

The area of the region is \(\frac{1024}{3}\). This corresponds to option (2).

Answer 2

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 \( 64 \): \[ \text{Area} = \int_{0}^{64} \left( 4\sqrt{x} - \frac{x}{2} \right) dx \] \[ \text{Area} = \int_{0}^{64} \left( 4x^{1/2} - \frac{1}{2}x \right) dx \] \[ \text{Area} = \left[ 4 \frac{x^{3/2}}{3/2} - \frac{1}{2} \frac{x^2}{2} \right]_{0}^{64} \] \[ \text{Area} = \left[ \frac{8}{3} x^{3/2} - \frac{1}{4} x^2 \right]_{0}^{64} \] \[ \text{Area} = \left( \frac{8}{3} (64)^{3/2} - \frac{1}{4} (64)^2 \right) - \left( \frac{8}{3} (0)^{3/2} - \frac{1}{4} (0)^2 \right) \] \[ \text{Area} = \frac{8}{3} (8^2)^{3/2} - \frac{1}{4} (4096) \] \[ \text{Area} = \frac{8}{3} (8^3) - 1024 \] \[ \text{Area} = \frac{8}{3} (512) - 1024 \] \[ \text{Area} = \frac{4096}{3} - \frac{3072}{3} \] \[ \text{Area} = \frac{1024}{3} \]