Sketch the region enclosed by the given curves and find its area using integration.  

y = 10 – x2 ; y = x2+ 2


To solve this problem, we need to follow these steps:

1. **Sketch the curves** \( y = 10 - x^2 \) and \( y = x^2 + 2 \).
2. **Find the points of intersection** by setting the equations equal to each other.
3. **Determine the limits of integration** from the points of intersection.
4. **Set up the integral** to find the area between the curves.
5. **Evaluate the integral** to find the area.

### 1. Sketch the Curves
- The curve \( y = 10 - x^2 \) is a downward-opening parabola with its vertex at \( (0, 10) \).
- The curve \( y = x^2 + 2 \) is an upward-opening parabola with its vertex at \( (0, 2) \).

### 2. Find Points of Intersection
To find the points of intersection, set the two equations equal to each other:
\[
10 - x^2 = x^2 + 2
\]

Simplify and solve for \( x \):
\[
10 - 2 = 2x^2
\]
\[
8 = 2x^2
\]
\[
x^2 = 4
\]
\[
x = \pm 2
\]

The points of intersection occur at \( x = -2 \) and \( x = 2 \). Plugging these back into either equation, the corresponding \( y \)-values are:
\[
y = (-2)^2 + 2 = 6 \quad \text{and} \quad y = (2)^2 + 2 = 6
\]

So, the points of intersection are \( (-2, 6) \) and \( (2, 6) \).

### 3. Set Up the Integral
The area between the curves is given by the integral of the difference between the upper function and the lower function. Here, \( y = 10 - x^2 \) is the upper function and \( y = x^2 + 2 \) is the lower function.

The area \( A \) is:
\[
A = \int_{-2}^{2} \left[ (10 - x^2) - (x^2 + 2) \right] \, dx
\]

Simplify the integrand:
\[
A = \int_{-2}^{2} \left( 10 - x^2 - x^2 - 2 \right) \, dx = \int_{-2}^{2} \left( 8 - 2x^2 \right) \, dx
\]

### 4. Evaluate the Integral
Now, integrate:
\[
A = \int_{-2}^{2} \left( 8 - 2x^2 \right) \, dx
\]
\[
A = \int_{-2}^{2} 8 \, dx - \int_{-2}^{2} 2x^2 \, dx
\]

Calculate each integral separately:
\[
\int_{-2}^{2} 8 \, dx = 8x \Big|_{-2}^{2} = 8(2) - 8(-2) = 16 + 16 = 32
\]
\[
\int_{-2}^{2} 2x^2 \, dx = 2\left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{2}{3} \left( 2^3 - (-2)^3 \right) = \frac{2}{3} \left( 8 - (-8) \right) = \frac{2}{3} \times 16 = \frac{32}{3}
\]

Thus, the area \( A \) is:
\[
A = 32 - \frac{32}{3} = \frac{96}{3} - \frac{32}{3} = \frac{64}{3}
\]

### Final Answer
The area enclosed by the curves is \( \frac{64}{3} \) square units.

Would you like more details on any part of this process?

### Related Questions

1. How would the area change if the curves intersected at different points?
2. What is the significance of symmetric limits in this integral?
3. How would you approach this problem if the curves were not parabolic?
4. What happens to the area if we change the equations to different parabolas?
5. Can you derive a general formula for the area between two parabolas?

### Tip
Always check the intersection points first to avoid integrating over incorrect limits.

Learn how to sketch the region enclosed by the curves y = 10 – x^2 and y = x^2 + 2 and find its area using integration. This problem involves advanced integration techniques and understanding of parabolic curves. Suitable for advanced high school students.

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