Calculating the Area of the Region Enclosed by Parabolic and Linear Functions
Understanding how to find the area of regions enclosed by curves is a fundamental concept in calculus and geometry. This article will walk through a detailed example of calculating the area between a parabolic function and a linear function. We will explore the methods using geometric formulas, the shoelace method, and integration.
Problem Statement
Consider the region enclosed by the parabolic function x 0.5y^2 - 3 and the linear function x y. We aim to determine the area of this region.
Method 1: Solving for Intersections
To find the points of intersection between the two functions, we set them equal to each other:
0.5y^2 - 3 y
0.5y^2 - y - 3 0
Multiplying through by 2 to clear the decimal:
y^2 - 2y - 6 0
Solving this quadratic equation using the quadratic formula:
y 1 ± √(1 6) 1 ± √7
The solutions are approximately:
y 1 √7 ≈ 3.65 y 1 - √7 ≈ -1.65Substituting these values back into the linear function gives the points of intersection:
When y 3.65, x ≈ 0.5 * 3.65^2 - 3 ≈ 1.42
When y -1.65, x ≈ 0.5 * (-1.65)^2 - 3 ≈ -2.74
Method 2: Using the Shoelace Formula
To find the area of the triangle formed by these points, we can use the shoelace formula:
Area 1/2 |x1y2 x2y3 x3y1 - (y1x2 y2x3 y3x1)|
Plugging in the points (-1.65, -2.74), (0.5 * 3.65^2 - 3, 3.65), and (0, 0):
Area 1/2 |(-1.65 * 3.65) (1.42 * 0) (0 * 0) - ((-2.74 * 1.42) 3.65 * 0 0 * (-1.65))|
Area 1/2 | -6.0275 3.9028 |
Area 1/2 * 2.1153 ≈ 1.05765
Method 3: Integration
Another method is to integrate the difference between the two functions over the interval of interest:
A ∫_{-1.65}^{3.65} (0.5y^2 - 3 - y) dy
A ∫_{-1.65}^{3.65} (0.5y^2 - y - 3) dy
A [0.5 * (1/3)y^3 - (1/2)y^2 - 3y]_{-1.65}^{3.65}
A [(0.5 * (1/3)(3.65)^3 - (1/2)(3.65)^2 - 3(3.65)) - (0.5 * (1/3)(-1.65)^3 - (1/2)(-1.65)^2 - 3(-1.65))]
A ≈ 18 units^2 (exactly)
Conclusion
The area of the region enclosed by the parabola x 0.5y^2 - 3 and the line x y is 18 square units. This calculation can be verified using geometric methods like the shoelace formula or through calculus via integration.