Calculating the Area Enclosed Between Two Parabolic Curves

Introduction

Finding the area enclosed between two curves can often be approached through integration, particularly when dealing with parabolic functions. This tutorial will walk you through the steps to calculate the enclosed area between the curves y2 4x and x2 4y. By following these detailed steps, you will not only solve the problem but also deepen your understanding of integration and curve properties.

Step 1: Determine the Points of Intersection

To begin our calculation, we need to find the points where the two given curves intersect. The given equations are: y2 4x x2 4y First, we rewrite the equations in terms of each other: x frac{y^2}{4} y frac{x^2}{4} By substituting the second equation into the first, we aim to find the x-values at the points of intersection. Let's proceed with this step:

Substituting y frac{x^2}{4} into y^2 4x, we get:

[ left(frac{x^2}{4}right)^2 4x ] Simplifying this equation, we obtain: [ frac{x^4}{16} 4x frac{x^4}{16} - 4x 0 frac{x(x^3 - 64)}{16} 0 ] This gives us two solutions: x 0 and x^3 64, from which we find x 4. Therefore, the x-values at the points of intersection are x 0 and x 4.

Now, we find the corresponding y-values for these x-values:

For x 0: [ y^2 4(0) implies y 0 text{The point is (0, 0).} ] For x 4: [ y^2 4(4) implies y pm 4 text{The points are (4, 4) and (4, -4).} ] Therefore, the points of intersection are (0, 0), (4, 4), and (4, -4).

Step 2: Set Up the Integral for Area

Next, we need to integrate the difference between the top and the bottom curves. Since we are dealing with a rightward-opening parabola and an upward-opening parabola, the top curve is y frac{x^2}{4} and the bottom curve is y 2sqrt{x} for the interval from x 0 to x 4.

Expressing this, we get:

[ A 2 int_{0}^{4} left(frac{x^2}{4} - 2sqrt{x}right)dx ]

Step 3: Calculate the Integral

Integrating the functions step-by-step:

First, we find the indefinite integral:

[ A 2 left[frac{x^3}{12} - frac{4x^{3/2}}{3}right]_0^4 ] Now, we evaluate the definite integral:

At x 4: [ frac{4^3}{12} - frac{4(4)^{3/2}}{3} frac{64}{12} - frac{32}{3} frac{64}{12} - frac{128}{12} -frac{64}{12} -frac{16}{3} ] At x 0, the value is 0.

Therefore, the area is:

[ A 2 left[-frac{16}{3} - 0right] -frac{32}{3} ] Since area cannot be negative, we take the absolute value:

The enclosed area is:

[ A frac{32}{3} ]

Conclusion

The area enclosed between the curves y^2 4x and x^2 4y is frac{32}{3} square units. This problem demonstrates the application of integration in finding the area between two parabolic curves, emphasizing the importance of determining points of intersection and carefully setting up the integral.