Calculating Volume Of Solid Of Revolution And Area Of Bounded Region

Introduction

In this comprehensive guide, we will delve into the fascinating world of calculus and explore how to calculate the volume of a solid of revolution and the area of a region bounded by curves. These concepts are fundamental in various fields, including engineering, physics, and computer graphics. We will tackle a specific problem involving the rotation of a region bounded by two curves around a vertical line, providing a step-by-step solution and insightful explanations. By the end of this article, you will have a solid understanding of the techniques and principles involved in solving such problems.

Problem Statement

The core of our discussion revolves around a specific challenge: finding the volume of the solid that is formed when the region enclosed by the curves x² - 3y = 3 and x² + 4y = 0 is rotated about the vertical line x = -7. This problem elegantly combines concepts from both algebra and integral calculus, requiring us to first identify the region of interest and then apply appropriate integration techniques to determine the volume of the resulting solid. This type of problem is a classic example of applications of calculus in real-world scenarios, where understanding volumes and areas of complex shapes is crucial. We will break down the problem into manageable steps, making the process clear and accessible even for those who are relatively new to these concepts. Our goal is not just to provide the solution, but also to explain the underlying principles and reasoning, so that you can apply these techniques to other similar problems.

1. Finding the Intersection Points

The initial step in solving this problem is to determine the points of intersection between the two curves. These points define the boundaries of the region that we will be rotating. To find these points, we need to solve the system of equations:

1. x² - 3y = 3
2. x² + 4y = 0

We can use a variety of methods to solve this system, including substitution or elimination. Let's use the elimination method. Subtracting the first equation from the second equation, we get:

( x² + 4y ) - ( x² - 3y ) = 0 - 3

Simplifying this equation, we have:

7y = -3

Solving for y, we find:

y = -3/7

Now, we can substitute this value of y into either of the original equations to solve for x. Let's use the second equation:

x² + 4(-3/7) = 0

x² - 12/7 = 0

x² = 12/7

Taking the square root of both sides, we get:

x = ±√(12/7) = ±2√(3/7)

Thus, the points of intersection are (2√(3/7), -3/7) and (-2√(3/7), -3/7). These points are critical because they define the limits of integration when we calculate the area and the volume of the solid of revolution. Accurately determining these points is a fundamental step in the problem-solving process. Understanding how the curves intersect helps us visualize the region that will be rotated, which is essential for setting up the integral correctly. The x-coordinates of these points, ±2√(3/7), will serve as the bounds of integration when we apply the methods of calculus to find the volume. This careful attention to detail in finding the intersection points is a hallmark of a strong mathematical approach.

2. Expressing y in Terms of x

To calculate the volume of the solid of revolution, we will use the method of cylindrical shells. This method requires us to express y in terms of x for both curves. From the given equations, we can isolate y as follows:

1. From x² - 3y = 3, we get: 3y = x² - 3, so y = (x² - 3)/3
2. From x² + 4y = 0, we get: 4y = -x², so y = -x²/4

These expressions for y as functions of x are crucial for setting up the integral that will give us the volume of the solid. When using the method of cylindrical shells, we integrate with respect to x, and the height of each cylindrical shell is the difference between the y-values of the two curves at a given x-value. Therefore, we need to have these functions in the form y = f(x) for both curves. The expressions we've derived, y = (x² - 3)/3 and y = -x²/4, allow us to determine the height of each cylindrical shell as a function of x. This step is a key preparatory step for applying the integral calculus techniques needed to find the volume. Understanding how to manipulate algebraic equations to isolate variables is a fundamental skill in calculus and is essential for solving problems involving areas and volumes.

3. Setting Up the Integral for Volume

Now that we have the points of intersection and the expressions for y in terms of x, we can set up the integral to calculate the volume of the solid. Since we are rotating the region about the line x = -7, we will use the method of cylindrical shells. The radius of each cylindrical shell is the distance from the line of rotation (x = -7) to a given x-value, which is x + 7. The height of each shell is the difference between the y-values of the two curves, which is (-x²/4) - ((x² - 3)/3). The thickness of each shell is dx. Therefore, the volume dV of a single cylindrical shell is given by:

dV = 2π(radius)(height)(thickness) = 2π(x + 7)[(-x²/4) - ((x² - 3)/3)]dx

The total volume V is then the integral of dV over the interval defined by the x-coordinates of the intersection points, which we found earlier to be -2√(3/7) and 2√(3/7). Thus, the integral for the volume is:

V = ∫_-2√(3/7)^2√(3/7) 2π(x + 7)[(-x²/4) - ((x² - 3)/3)]dx

This integral represents the sum of the volumes of infinitesimally thin cylindrical shells that make up the solid of revolution. The setup of this integral is a critical step in the problem-solving process, as it translates the geometric concept of volume into a mathematical expression that can be evaluated. Understanding the method of cylindrical shells and how it relates to the geometry of the solid is essential for setting up the integral correctly. The limits of integration are determined by the intersection points of the curves, and the integrand represents the volume of a single cylindrical shell. This step requires a deep understanding of the principles of integral calculus and the geometry of solids of revolution.

4. Evaluating the Integral

The next step is to evaluate the integral we set up in the previous section. This involves simplifying the integrand and then applying the fundamental theorem of calculus. Let's first simplify the expression inside the integral:

(-x²/4) - ((x² - 3)/3) = (-3x² - 4x² + 12)/12 = (-7x² + 12)/12

So the integral becomes:

V = 2π ∫_-2√(3/7)^2√(3/7) (x + 7)((-7x² + 12)/12) dx

Now, let's expand the integrand:

(x + 7)((-7x² + 12)/12) = (-7x³ + 12x - 49x*² + 84)/12

Thus, the integral is:

V = (π/6) ∫_-2√(3/7)^2√(3/7) (-7x³ - 49x² + 12*x + 84) dx

We can now integrate term by term:

V = (π/6) [-7x⁴/4 - 49x³/3 + 6x² + 84*x]_-2√(3/7)^2√(3/7)

Evaluating the integral at the limits of integration, we get:

V = (π/6) [(-7/4)(2√(3/7))⁴ - (49/3)(2√(3/7))³ + 6(2√(3/7))² + 84(2√(3/7)) - (-7/4)(-2√(3/7))⁴ + (49/3)(-2√(3/7))³ - 6(-2√(3/7))² - 84(-2√(3/7))]

Simplifying this expression, we find:

V = (π/6) [0 - (98/3)(2√(3/7))³ + 0 + 168(2√(3/7))]

V = (π/6) [-(98/3)(2√(3/7))³ + 168(2√(3/7))]

V ≈ 147.65

Therefore, the volume of the solid formed when the region is rotated about the line x = -7 is approximately 147.65 cubic units. This process of evaluating the definite integral involves a series of algebraic manipulations and the application of the power rule for integration. The accurate evaluation of the integral is crucial for obtaining the correct answer. This step highlights the importance of both algebraic skills and a thorough understanding of the fundamental theorem of calculus. The final numerical result provides a quantitative measure of the volume of the solid, demonstrating the power of calculus in solving geometric problems.

5. Calculating the Area of the Region

While the primary focus of the problem is on finding the volume, let's also calculate the area of the region bounded by the curves x² - 3y = 3 and x² + 4y = 0. This will provide a more complete understanding of the region we are working with. The area A of the region can be found by integrating the difference between the y-values of the two curves with respect to x over the interval defined by the points of intersection. We already found the expressions for y in terms of x:

1. y₁ = -x²/4
2. y₂ = (x² - 3)/3

The area A is given by the integral:

A = ∫_-2√(3/7)^2√(3/7) [ y₁ - y₂ ] dx = ∫_-2√(3/7)^2√(3/7) [ (-x²/4) - ((x² - 3)/3) ] dx

We already simplified the integrand in the volume calculation:

(-x²/4) - ((x² - 3)/3) = (-7x² + 12)/12

So the integral for the area is:

A = ∫_-2√(3/7)^2√(3/7) (-7x² + 12)/12 dx = (1/12) ∫_-2√(3/7)^2√(3/7) (-7x² + 12) dx

Now, we integrate term by term:

A = (1/12) [-7x³/3 + 12*x]_-2√(3/7)^2√(3/7)

Evaluating the integral at the limits of integration, we get:

A = (1/12) [(-7/3)(2√(3/7))³ + 12(2√(3/7)) - (-7/3)(-2√(3/7))³ + 12(-2√(3/7))]

Simplifying this expression, we find:

A = (1/12) [(-14/3)(2√(3/7))³ + 24(2√(3/7))]

A ≈ 4.37

Thus, the area of the region bounded by the curves is approximately 4.37 square units. Calculating the area of the region provides a valuable context for understanding the size and shape of the region that was rotated to form the solid. This calculation reinforces the principles of integral calculus and demonstrates how to apply them to find areas bounded by curves. The result gives a quantitative measure of the region's size, which complements the volume calculation and provides a more complete picture of the problem.

Conclusion

In this article, we have successfully calculated the volume of the solid of revolution formed by rotating the region bounded by the curves x² - 3y = 3 and x² + 4y = 0 about the line x = -7. We used the method of cylindrical shells, which involved finding the points of intersection, expressing y in terms of x, setting up the integral, and evaluating it. The calculated volume is approximately 147.65 cubic units. Additionally, we found the area of the region to be approximately 4.37 square units. This comprehensive solution demonstrates the power of calculus in solving geometric problems. Understanding these techniques is essential for anyone studying engineering, physics, or other related fields. The process we followed highlights the importance of breaking down complex problems into smaller, manageable steps. From finding intersection points to setting up and evaluating integrals, each step requires a solid understanding of the underlying principles of calculus. By mastering these techniques, you can tackle a wide range of problems involving areas, volumes, and other geometric quantities. The combination of algebraic manipulation and integral calculus is a powerful tool for solving real-world problems, and this article provides a solid foundation for further exploration of these concepts.