If the area of the region $\{(x,y):|4-x^{2}|\le y\le x^{2},y\le4,x\ge0\}$ is $(\frac{80\sqrt{2}}{\alpha}-\beta),\alpha, \beta\in \mathbb{N}$ then $\alpha+\beta$ is equal to:
Solution:
The region is defined by the following inequalities:
1. $y \ge |4-x^2|$
2. $y \le x^2$
3. $y \le 4$
4. $x \ge 0$
First, determine the range of $x$ where the area exists by finding the intersection points of the boundary curves.
The condition $|4-x^2| \le x^2$ implies the lower boundary is below the upper boundary.
Case 1: $x^2 \le 4$ (i.e., $0 \le x \le 2$)
$$ 4-x^2 \le x^2 \Rightarrow 2x^2 \ge 4 \Rightarrow x^2 \ge 2 \Rightarrow x \ge \sqrt{2} $$
So, the valid interval is $[\sqrt{2}, 2]$.
In this interval, the region is bounded below by $y = 4-x^2$ and above by $y = x^2$.
Also, $y \le 4$ is satisfied since $x^2 \le 4$.
Case 2: $x^2 > 4$ (i.e., $x > 2$)
$$ -(4-x^2) \le x^2 \Rightarrow x^2-4 \le x^2 \Rightarrow -4 \le 0 $$
This is always true. However, we also have the constraint $y \le 4$.
The upper bound is now $y=4$ (since $x^2 > 4$).
The lower bound is $y = x^2-4$ (since $|4-x^2| = x^2-4$ for $x>2$).
We need to find where these intersect:
$$ x^2-4 \le 4 \Rightarrow x^2 \le 8 \Rightarrow x \le 2\sqrt{2} $$
So, the valid interval is $[2, 2\sqrt{2}]$.
Now, set up the integrals for the two regions:
$$ A = \int_{\sqrt{2}}^{2} (x^2 – (4-x^2)) dx + \int_{2}^{2\sqrt{2}} (4 – (x^2-4)) dx $$
$$ A = \int_{\sqrt{2}}^{2} (2x^2 – 4) dx + \int_{2}^{2\sqrt{2}} (8 – x^2) dx $$
Calculate the first integral:
$$ \left[ \frac{2x^3}{3} – 4x \right]_{\sqrt{2}}^{2} = \left( \frac{16}{3} – 8 \right) – \left( \frac{4\sqrt{2}}{3} – 4\sqrt{2} \right) $$
$$ = \left( -\frac{8}{3} \right) – \left( -\frac{8\sqrt{2}}{3} \right) = \frac{8\sqrt{2} – 8}{3} $$
Calculate the second integral:
$$ \left[ 8x – \frac{x^3}{3} \right]_{2}^{2\sqrt{2}} = \left( 16\sqrt{2} – \frac{16\sqrt{2}}{3} \right) – \left( 16 – \frac{8}{3} \right) $$
$$ = \left( \frac{32\sqrt{2}}{3} \right) – \left( \frac{40}{3} \right) = \frac{32\sqrt{2} – 40}{3} $$
Add both areas:
$$ A = \frac{8\sqrt{2} – 8}{3} + \frac{32\sqrt{2} – 40}{3} = \frac{40\sqrt{2} – 48}{3} $$
$$ A = \frac{40\sqrt{2}}{3} – 16 $$
We need to match the form $\frac{80\sqrt{2}}{\alpha} – \beta$.
Multiply numerator and denominator of the first term by 2:
$$ A = \frac{80\sqrt{2}}{6} – 16 $$
Comparing with the given form:
$$ \alpha = 6, \quad \beta = 16 $$
$$ \alpha + \beta = 6 + 16 = 22 $$
Ans. (22)