Question

If the area of the region \[ \{(x, y) : |4 - x^2| \leq y \leq x^2, y \leq 4, x \geq 0\} \] is \( \frac{80\sqrt{2}}{\alpha - \beta} \), where \( \alpha, \beta \in \mathbb{N} \), then \( \alpha + \beta \) is equal to:

Hint:

For integration of absolute values and square roots, break the function into intervals and evaluate using standard methods.

Answer 1

Correct Answer: 22

We are asked to find the value of \( \alpha + \beta \) given that the area of the region defined by \( \{(x, y) : |4 - x^2| \le y \le x^2, y \le 4, x \ge 0\} \) is equal to \( \left( \frac{80\sqrt{2}}{\alpha} - \beta \right) \), where \( \alpha, \beta \in \mathbb{N} \).

Concept Used:

The area of a region bounded by curves can be found using definite integration. The area between two curves \( y = f(x) \) (upper curve) and \( y = g(x) \) (lower curve) from \( x = a \) to \( x = b \) is given by:

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

We must first analyze the inequalities, especially the one involving the absolute value, to determine the exact boundaries and the intervals of integration.

Step-by-Step Solution:

Step 1: Analyze the inequalities to define the region.

The region is defined by the following conditions:

1. \( |4 - x^2| \le y \le x^2 \)
2. \( y \le 4 \)
3. \( x \ge 0 \)

From the first inequality, we must have \( |4 - x^2| \le x^2 \). This can be written as:

\[ -x^2 \le 4 - x^2 \le x^2 \]

This gives two separate inequalities:

• \( -x^2 \le 4 - x^2 \implies 0 \le 4 \), which is always true.
• \( 4 - x^2 \le x^2 \implies 4 \le 2x^2 \implies x^2 \ge 2 \).

Since we are given \( x \ge 0 \), the condition \( x^2 \ge 2 \) implies \( x \ge \sqrt{2} \). So, the region only exists for \( x \ge \sqrt{2} \).

Step 2: Split the region of integration based on the absolute value function.

The expression inside the absolute value, \( 4 - x^2 \), changes sign at \( x = 2 \). We will split the integration at this point.

Case 1: \( \sqrt{2} \le x \le 2 \)

In this interval, \( 2 \le x^2 \le 4 \), so \( 4 - x^2 \ge 0 \). The inequality \( |4 - x^2| \le y \) becomes \( 4 - x^2 \le y \).

The bounds on \( y \) are \( 4 - x^2 \le y \le x^2 \). The condition \( y \le 4 \) is automatically satisfied since \( y \le x^2 \le 4 \).

Case 2: \( x > 2 \)

In this interval, \( x^2 > 4 \), so \( 4 - x^2 < 0 \). The inequality \( |4 - x^2| \le y \) becomes \( -(4 - x^2) \le y \), or \( x^2 - 4 \le y \).

The bounds on \( y \) are \( x^2 - 4 \le y \le x^2 \). We also have the condition \( y \le 4 \). This means the upper bound is \( \min(x^2, 4) \). Since \( x > 2 \), \( x^2 > 4 \), so the upper bound is \( y = 4 \).

The bounds become \( x^2 - 4 \le y \le 4 \). For a region to exist, we must have \( x^2 - 4 \le 4 \), which means \( x^2 \le 8 \), or \( x \le \sqrt{8} = 2\sqrt{2} \). So this case applies for \( 2 < x \le 2\sqrt{2} \).

Step 3: Set up and calculate the integral for the first part of the area (\( A_1 \)).

For the interval \( [\sqrt{2}, 2] \), the area is:

\[ A_1 = \int_{\sqrt{2}}^{2} (x^2 - (4 - x^2)) \,dx = \int_{\sqrt{2}}^{2} (2x^2 - 4) \,dx \] \[ = \left[ \frac{2x^3}{3} - 4x \right]_{\sqrt{2}}^{2} = \left(\frac{2(2)^3}{3} - 4(2)\right) - \left(\frac{2(\sqrt{2})^3}{3} - 4\sqrt{2}\right) \] \[ = \left(\frac{16}{3} - 8\right) - \left(\frac{4\sqrt{2}}{3} - 4\sqrt{2}\right) = \left(\frac{16-24}{3}\right) - \left(\frac{4\sqrt{2}-12\sqrt{2}}{3}\right) \] \[ = -\frac{8}{3} - \left(-\frac{8\sqrt{2}}{3}\right) = \frac{8\sqrt{2}}{3} - \frac{8}{3} \]

Step 4: Set up and calculate the integral for the second part of the area (\( A_2 \)).

For the interval \( (2, 2\sqrt{2}] \), the area is:

\[ A_2 = \int_{2}^{2\sqrt{2}} (4 - (x^2 - 4)) \,dx = \int_{2}^{2\sqrt{2}} (8 - x^2) \,dx \] \[ = \left[ 8x - \frac{x^3}{3} \right]_{2}^{2\sqrt{2}} = \left(8(2\sqrt{2}) - \frac{(2\sqrt{2})^3}{3}\right) - \left(8(2) - \frac{2^3}{3}\right) \] \[ = \left(16\sqrt{2} - \frac{16\sqrt{2}}{3}\right) - \left(16 - \frac{8}{3}\right) = \left(\frac{48\sqrt{2}-16\sqrt{2}}{3}\right) - \left(\frac{48-8}{3}\right) \] \[ = \frac{32\sqrt{2}}{3} - \frac{40}{3} \]

Step 5: Calculate the total area.

Total Area \( A = A_1 + A_2 \):

\[ A = \left(\frac{8\sqrt{2}}{3} - \frac{8}{3}\right) + \left(\frac{32\sqrt{2}}{3} - \frac{40}{3}\right) = \frac{40\sqrt{2}}{3} - \frac{48}{3} = \frac{40\sqrt{2}}{3} - 16 \]

Final Computation & Result:

We are given that the area is \( \left( \frac{80\sqrt{2}}{\alpha} - \beta \right) \). We equate this with our calculated area.

\[ \frac{80\sqrt{2}}{\alpha} - \beta = \frac{40\sqrt{2}}{3} - 16 \]

To match the form, we can write our result as:

\[ \frac{40\sqrt{2} \times 2}{3 \times 2} - 16 = \frac{80\sqrt{2}}{6} - 16 \]

By comparing the two expressions, we can identify \( \alpha \) and \( \beta \):

\[ \alpha = 6 \quad \text{and} \quad \beta = 16 \]

Both \( \alpha \) and \( \beta \) are natural numbers, as required. The question asks for the value of \( \alpha + \beta \).

\[ \alpha + \beta = 6 + 16 = 22 \]

The value of \( \alpha + \beta \) is 22.