1. Given region

We need the area of

$$\{(x,y): |4-x^2|\le y\le x^2,\ y\le 4,\ x\ge 0\}.$$

So for each $x\ge 0$, the allowed $y$ must satisfy:

• lower bound: $y\ge |4-x^2|$
• upper bounds: $y\le x^2$ and $y\le 4$

Hence the actual upper bound is

$$y\le \min(x^2,4).$$

So the vertical strip exists when

$$|4-x^2|\le \min(x^2,4).$$

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2. Split according to $x$

Because of $x^2$ and $4$, the natural breakpoints are around $x=2$.

Case 1: $0\le x\le 2$

Then $x^2\le 4$, so

$$\min(x^2,4)=x^2.$$

Also,

$$|4-x^2|=4-x^2$$

since $x^2\le 4$.

Thus the strip is

$$4-x^2\le y\le x^2.$$

This is possible only if

$$4-x^2\le x^2 \implies 4\le 2x^2 \implies x^2\ge 2 \implies x\ge \sqrt{2}.$$

So in this range, valid $x$ are

$$\sqrt{2}\le x\le 2.$$

The height of the strip is

$$x^2-(4-x^2)=2x^2-4.$$

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Case 2: $x\ge 2$

Then $x^2\ge 4$, so

$$\min(x^2,4)=4.$$

Also,

$$|4-x^2|=x^2-4.$$

Thus the strip is

$$x^2-4\le y\le 4.$$

This is possible only if

$$x^2-4\le 4 \implies x^2\le 8 \implies x\le 2\sqrt{2}.$$

So in this range, valid $x$ are

$$2\le x\le 2\sqrt{2}.$$

The height of the strip is

$$4-(x^2-4)=8-x^2.$$

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3. Set up the area integral

Therefore,

$$A=\int_{\sqrt{2}}^{2}(2x^2-4)\,dx+\int_{2}^{2\sqrt{2}}(8-x^2)\,dx.$$

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4. Evaluate the first integral

$$\int (2x^2-4)dx=\frac{2x^3}{3}-4x.$$

So

$$\int_{\sqrt{2}}^{2}(2x^2-4)dx =\left[\frac{2x^3}{3}-4x\right]_{\sqrt{2}}^{2}.$$

At $x=2$:

$$\frac{2(8)}{3}-8=\frac{16}{3}-8=-\frac{8}{3}.$$

At $x=\sqrt{2}$:

$$\frac{2(\sqrt{2})^3}{3}-4\sqrt{2} =\frac{2(2\sqrt{2})}{3}-4\sqrt{2} =\frac{4\sqrt{2}}{3}-4\sqrt{2} =-\frac{8\sqrt{2}}{3}.$$

Therefore,

$$I_1=-\frac{8}{3}-\left(-\frac{8\sqrt{2}}{3}\right) =\frac{8\sqrt{2}-8}{3}.$$

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5. Evaluate the second integral

$$\int (8-x^2)dx=8x-\frac{x^3}{3}.$$

So

$$\int_{2}^{2\sqrt{2}}(8-x^2)dx =\left[8x-\frac{x^3}{3}\right]_{2}^{2\sqrt{2}}.$$

At $x=2\sqrt{2}$:

$$8(2\sqrt{2})-\frac{(2\sqrt{2})^3}{3} =16\sqrt{2}-\frac{16\sqrt{2}}{3} =\frac{32\sqrt{2}}{3}.$$

At $x=2$:

$$16-\frac{8}{3}=\frac{40}{3}.$$

Therefore,

$$I_2=\frac{32\sqrt{2}}{3}-\frac{40}{3}.$$

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6. Total area

$$A=I_1+I_2 =\frac{8\sqrt{2}-8}{3}+\frac{32\sqrt{2}-40}{3} =\frac{40\sqrt{2}-48}{3}.$$

So

$$A=\frac{40\sqrt{2}}{3}-16.$$

This matches the form

$$\frac{80\sqrt{2}}{\alpha}-\beta.$$

Thus,

$$\frac{80\sqrt{2}}{\alpha}=\frac{40\sqrt{2}}{3} \implies \alpha=6,$$

and

$$\beta=16.$$

Hence,

$$\alpha+\beta=6+16=22.$$

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7. Comparison with stored answer

Derived answer = $22$. Stored correct answer = $22$.

They agree.