If the area of the region (x,y):|4-x^2| y x^2,y 4,x 0 is (80√2 - ), , N, then + is equal t…

Question

If the area of the region ${(x,y):|4-{x}^{2}|\le y\le {x}^{2},y\le 4,x\ge 0}$ is $(\frac{80\sqrt{2}}{\alpha }-\beta ),\alpha ,\beta \in N$, then $\alpha +\beta $ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 22

Accepted Answer

For $0\le x\le\sqrt2$, $|4-x^2|=4-x^2$ (upper bound uses this branch); for $\sqrt2\le x\le2$, $|4-x^2|=4-x^2$ still (since $x^2\le4$) but the lower boundary flips relative to $x^2$. Splitting the integral at $x=\sqrt2$ and $x=2$, and combining with $y\le x^2$, $y\le4$ appropriately (as in the source computation):

$$\text{Area} = \frac{40\sqrt2}{3}-16$$

Matching to $\dfrac{80\sqrt2}{\alpha}-\beta$: $\alpha=6$, $\beta=16$, so $\alpha+\beta = 22$.

Solutions · 1