Math Problem Statement

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

y = 10 – x2 ; y = x2+ 2

Solution

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

  1. Sketch the curves y=10x2y = 10 - x^2 and y=x2+2y = 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=10x2y = 10 - x^2 is a downward-opening parabola with its vertex at (0,10)(0, 10).
  • The curve y=x2+2y = x^2 + 2 is an upward-opening parabola with its vertex at (0,2)(0, 2).

2. Find Points of Intersection

To find the points of intersection, set the two equations equal to each other: 10x2=x2+210 - x^2 = x^2 + 2

Simplify and solve for xx: 102=2x210 - 2 = 2x^2 8=2x28 = 2x^2 x2=4x^2 = 4 x=±2x = \pm 2

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

So, the points of intersection are (2,6)(-2, 6) and (2,6)(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=10x2y = 10 - x^2 is the upper function and y=x2+2y = x^2 + 2 is the lower function.

The area AA is: A=22[(10x2)(x2+2)]dxA = \int_{-2}^{2} \left[ (10 - x^2) - (x^2 + 2) \right] \, dx

Simplify the integrand: A=22(10x2x22)dx=22(82x2)dxA = \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=22(82x2)dxA = \int_{-2}^{2} \left( 8 - 2x^2 \right) \, dx A=228dx222x2dxA = \int_{-2}^{2} 8 \, dx - \int_{-2}^{2} 2x^2 \, dx

Calculate each integral separately: 228dx=8x22=8(2)8(2)=16+16=32\int_{-2}^{2} 8 \, dx = 8x \Big|_{-2}^{2} = 8(2) - 8(-2) = 16 + 16 = 32 222x2dx=2[x33]22=23(23(2)3)=23(8(8))=23×16=323\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 AA is: A=32323=963323=643A = 32 - \frac{32}{3} = \frac{96}{3} - \frac{32}{3} = \frac{64}{3}

Final Answer

The area enclosed by the curves is 643\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.

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Math Problem Analysis

Mathematical Concepts

Integration
Area between curves
Parabolas

Formulas

Integral of a difference of functions

Theorems

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Suitable Grade Level

Advanced High School