If the area of the region $\{(x, y):|4-x^2|\le y \le x^2, y\le4,x\ge0)$ is $\left(\frac{80\sqrt{2}}{\alpha}-\beta\right)$, $\alpha, \beta\in N$, then $\alpha+\beta$ is equal to _____________.
Calculating Area of Region Defined by Inequalities and Finding Alpha + Beta
The problem requires finding the area of a region defined by the inequalities $|4-x^2| \le y \le x^2$, $y \le 4$, and $x \ge 0$. The area is given in the format $\left(\frac{80\sqrt{2}}{\alpha}-\beta\right)$, where $\alpha, \beta$ are natural numbers. We need to find the value of $\alpha + \beta$.
Understanding the Region Boundaries
First, let's analyze the absolute value inequality $|4-x^2|$. We consider two cases based on the sign of $4-x^2$:
- Case 1: $4-x^2 \ge 0$
This occurs when $x^2 \le 4$. Since $x \ge 0$, this means $0 \le x \le 2$. In this interval, $|4-x^2| = 4-x^2$. The inequalities become $4-x^2 \le y \le x^2$ and $y \le 4$.
The condition $4-x^2 \le y \le x^2$ implies $4-x^2 \le x^2$, which simplifies to $4 \le 2x^2$, or $x^2 \ge 2$. Since $x \ge 0$, this means $x \ge \sqrt{2}$. Thus, this case applies for $\sqrt{2} \le x \le 2$. The upper bound $y \le 4$ is automatically satisfied because $y \le x^2$ and $x \le 2$ means $x^2 \le 4$. The region is defined by $\sqrt{2} \le x \le 2$ and $4-x^2 \le y \le x^2$.
- Case 2: $4-x^2 < 0$
This occurs when $x^2 > 4$. Since $x \ge 0$, this means $x > 2$. In this interval, $|4-x^2| = -(4-x^2) = x^2-4$. The inequalities become $x^2-4 \le y \le x^2$ and $y \le 4$.
Combining the upper bounds $y \le x^2$ and $y \le 4$, we get $y \le \min(x^2, 4)$. Since $x > 2$, $x^2 > 4$, so $\min(x^2, 4) = 4$. The inequalities simplify to $x^2-4 \le y \le 4$. For this region to exist, we must have $x^2-4 \le 4$, which means $x^2 \le 8$. Since $x > 2$, this case applies for $2 < x \le \sqrt{8} = 2\sqrt{2}$. The region is defined by $2 < x \le 2\sqrt{2}$ and $x^2-4 \le y \le 4$.
Calculating the Area
The total area is the sum of the areas from the two cases.
Area from Case 1 ($\sqrt{2} \le x \le 2$):
$A_1 = \int_{\sqrt{2}}^{2} (x^2 - (4-x^2)) dx = \int_{\sqrt{2}}^{2} (2x^2 - 4) dx$ $A_1 = \left[ \frac{2x^3}{3} - 4x \right]_{\sqrt{2}}^{2}$ $A_1 = \left( \frac{2(2)^3}{3} - 4(2) \right) - \left( \frac{2(\sqrt{2})^3}{3} - 4\sqrt{2} \right)$ $A_1 = \left( \frac{16}{3} - 8 \right) - \left( \frac{4\sqrt{2}}{3} - 4\sqrt{2} \right)$ $A_1 = \left( \frac{16 - 24}{3} \right) - \left( \frac{4\sqrt{2} - 12\sqrt{2}}{3} \right)$ $A_1 = -\frac{8}{3} - \left( -\frac{8\sqrt{2}}{3} \right) = \frac{8\sqrt{2} - 8}{3}$Area from Case 2 ($2 < x \le 2\sqrt{2}$):
$A_2 = \int_{2}^{2\sqrt{2}} (4 - (x^2-4)) dx = \int_{2}^{2\sqrt{2}} (8 - x^2) dx$ $A_2 = \left[ 8x - \frac{x^3}{3} \right]_{2}^{2\sqrt{2}}$ $A_2 = \left( 8(2\sqrt{2}) - \frac{(2\sqrt{2})^3}{3} \right) - \left( 8(2) - \frac{2^3}{3} \right)$ $A_2 = \left( 16\sqrt{2} - \frac{16\sqrt{2}}{3} \right) - \left( 16 - \frac{8}{3} \right)$ $A_2 = \left( \frac{48\sqrt{2} - 16\sqrt{2}}{3} \right) - \left( \frac{48 - 8}{3} \right)$ $A_2 = \frac{32\sqrt{2}}{3} - \frac{40}{3} = \frac{32\sqrt{2} - 40}{3}$Total Area:
$A = A_1 + A_2 = \frac{8\sqrt{2} - 8}{3} + \frac{32\sqrt{2} - 40}{3}$ $A = \frac{(8\sqrt{2} + 32\sqrt{2}) - (8 + 40)}{3} = \frac{40\sqrt{2} - 48}{3}$ $A = \frac{40\sqrt{2}}{3} - 16$Finding Alpha and Beta
The calculated area is $A = \frac{40\sqrt{2}}{3} - 16$. We are given the area in the form $\left(\frac{80\sqrt{2}}{\alpha}-\beta\right)$.
Comparing the two forms:
- For the term with $\sqrt{2}$: $\frac{40\sqrt{2}}{3} = \frac{80\sqrt{2}}{\alpha}$. Multiplying both sides by $\frac{3\alpha}{40\sqrt{2}}$ gives $\alpha = \frac{80 \times 3}{40} = 2 \times 3 = 6$.
- For the constant term: $\beta = 16$.
Both $\alpha = 6$ and $\beta = 16$ are natural numbers, satisfying the condition $\alpha, \beta \in N$.
Calculating Alpha + Beta
The final step is to calculate $\alpha + \beta$.
$\alpha + \beta = 6 + 16 = 22$Similar Questions
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