To find the area of the enclosed region between the graphs of $y = 3$ and $y = -x^2 + 7$, we first need to determine the points of intersection between these two curves.

Step 1: Solve for points of intersection

The curves intersect where $-x^2 + 7 = 3$. Simplify the equation: $-x^2 + 7 = 3 \implies -x^2 = -4 \implies x^2 = 4.$ Thus, $x = \pm 2$.

The points of intersection are $(-2, 3)$ and $(2, 3)$.

Step 2: Set up the integral

The area is determined by integrating the difference between the upper curve ($y = -x^2 + 7$) and the lower curve ($y = 3$) from $x = -2$ to $x = 2$: $\text{Area} = \int_{-2}^2 \left[(-x^2 + 7) - 3\right] dx.$

Simplify the integrand: $(-x^2 + 7) - 3 = -x^2 + 4.$ Thus, the area is: $\text{Area} = \int_{-2}^2 (-x^2 + 4) dx.$

Step 3: Compute the integral

Break it down: $\int_{-2}^2 (-x^2 + 4) dx = \int_{-2}^2 -x^2 dx + \int_{-2}^2 4 dx.$

(a) Compute $\int_{-2}^2 -x^2 dx$:

Since $-x^2$ is an even function (symmetric about the $y$-axis), the integral over symmetric limits cancels out: $\int_{-2}^2 -x^2 dx = 0.$

(b) Compute $\int_{-2}^2 4 dx$:

$\int_{-2}^2 4 dx = 4 \int_{-2}^2 1 dx = 4[x]_{-2}^2 = 4(2 - (-2)) = 4 \cdot 4 = 16.$

Step 4: Combine results

The total area is: $\text{Area} = 16.$

Final Answer:

The area of the enclosed region is: $\boxed{16 \, \text{square units}}$

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Would you like a step-by-step breakdown of integration or a graphical visualization?

Here are 5 related questions to consider:

1. How is the symmetry of the function used to simplify integration?
2. What changes if the enclosed region is bounded differently (e.g., different limits)?
3. How would the solution differ if we were finding the volume of a rotated region?
4. What if the lower function were shifted vertically, e.g., $y = -x^2 + 5$?
5. How can we use numerical methods to verify this integral?

Tip: Always sketch the curves to visually confirm the region you're integrating!