Sketch the region enclosed by the given curves on the given interval.



*y* = 6 − 3*x*2,    *y* = 3*x*2,         −3 ≤ *x* ≤ 3

Find the area of the region.

Here is the sketch of the region enclosed by the curves \( y = 6 - 3x^2 \) (blue parabola) and \( y = 3x^2 \) (red parabola) on the interval \( -3 \leq x \leq 3 \). The shaded purple area represents the enclosed region.

The area of the enclosed region between \( x = -1 \) and \( x = 1 \) is \( 8 \) square units.

Would you like more details or have any further questions?

Here are five related questions you might find interesting:
1. How would the area change if the interval was expanded to \( -2 \leq x \leq 2 \)?
2. What are the key differences between finding areas under curves and between curves?
3. Can you derive the equation of the axis of symmetry for both parabolas?
4. What happens to the enclosed region if the downward parabola is shifted up or down?
5. How would you apply this method to find the volume of the region when rotated about the x-axis?

**Tip:** When finding the area between curves, always subtract the lower curve from the upper curve in your integral.

Explore how to calculate the area enclosed by the parabolas y = 6 - 3x^2 (blue) and y = 3x^2 (red) on the interval [-3, 3]. This problem involves finding the area between curves, utilizing calculus and parabolic equations. Suitable for Grades 11-12.

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