1. Understand the region

We need the area of

$$\{(x,y): x^2 \le y \le 8-x^2,\; y\le 7\}.$$

This means:

• lower boundary: $y=x^2$
• upper boundary: $y=8-x^2$
• also restricted by $y\le 7$

So the actual upper boundary is the smaller of $7$ and $8-x^2$.

Hence the region is bounded by

$$y=x^2 \quad \text{and} \quad y=\min(7,\,8-x^2).$$

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2. Find where the top boundary changes

Compare $7$ and $8-x^2$:

$$8-x^2 \ge 7 \iff x^2\le 1 \iff |x|\le 1.$$

So:

• for $|x|\le 1$, the upper boundary is $y=7$
• for $|x|>1$, the upper boundary is $y=8-x^2$

Also, for the region to exist, upper boundary must be at least lower boundary.

Using $y=x^2$ and $y=8-x^2$:

$$x^2 \le 8-x^2 \iff 2x^2\le 8 \iff x^2\le 4 \iff |x|\le 2.$$

Thus the region exists for $x\in[-2,2]$.

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3. Split the area into parts

Part 1: $-2\le x\le -1$

Here upper curve is $8-x^2$. Area contribution:

$$\int_{-2}^{-1}\big[(8-x^2)-x^2\big]dx =\int_{-2}^{-1}(8-2x^2)dx.$$

Part 2: $-1\le x\le 1$

Here upper curve is $7$. Area contribution:

$$\int_{-1}^{1}(7-x^2)dx.$$

Part 3: $1\le x\le 2$

Again upper curve is $8-x^2$. Area contribution:

$$\int_{1}^{2}(8-2x^2)dx.$$

By symmetry, Part 1 and Part 3 are equal. So

$$A=2\int_{1}^{2}(8-2x^2)dx+\int_{-1}^{1}(7-x^2)dx.$$

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4. Evaluate the integrals

First,

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

So,

$$\int_{1}^{2}(8-2x^2)dx =\left[8x-\frac{2x^3}{3}\right]_{1}^{2} =\left(16-\frac{16}{3}\right)-\left(8-\frac{2}{3}\right).$$

Compute:

$$16-\frac{16}{3}=\frac{48-16}{3}=\frac{32}{3},$$ $$8-\frac{2}{3}=\frac{24-2}{3}=\frac{22}{3}.$$

Thus,

$$\int_{1}^{2}(8-2x^2)dx=\frac{32}{3}-\frac{22}{3}=\frac{10}{3}.$$

Therefore the two side parts give

$$2\cdot \frac{10}{3}=\frac{20}{3}.$$

Now the middle part:

$$\int_{-1}^{1}(7-x^2)dx=\int_{-1}^{1}7\,dx-\int_{-1}^{1}x^2dx.$$

Compute:

$$\int_{-1}^{1}7\,dx=14,$$

and

$$\int_{-1}^{1}x^2dx=2\int_0^1 x^2dx=2\cdot\frac{1}{3}=\frac{2}{3}.$$

So,

$$\int_{-1}^{1}(7-x^2)dx=14-\frac{2}{3}=\frac{40}{3}.$$

Hence total area is

$$A=\frac{20}{3}+\frac{40}{3}=\frac{60}{3}=20.$$

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5. Identify the correct option

The area is

$$\boxed{20}.$$

So the correct option is C.