Problem 22 - Basic Technical Mathematics with Calculus

Question

Find the volume generated by revolving the regions bounded by the given curves about the \(y\) -axis. Use the indicated method in each case. $$y=3 x^{2}-x^{3}, y=0 \quad \text { (shells) }$$

Step-by-Step Solution

Answer

The volume is \(\frac{243\pi}{10}\).

Step 1: Set Up the Problem

We are given the curves \( y = 3x^2 - x^3 \) and \( y = 0 \) and asked to find the volume when these curves are revolved around the \( y \)-axis using the shell method. The shell method involves integrating cylindrical shells described by the given region.

Step 2: Identify Interval of Revolution

First, determine the bounds of the curves. Set \( y = 3x^2 - x^3 = 0 \) to find the \( x \)-intercepts: \( x(3x-x^2) = x(3-x)x = 0 \). The solutions are \( x = 0 \) and \( x = 3 \). Thus, the interval of revolution is from \( x = 0 \) to \( x = 3 \).

Step 3: Write the Expression for Cylindrical Shells

The formula for the volume of a shell is given by \( V = 2\pi \int_{a}^{b} p(x) h(x) \, dx \), where \( p(x) \) is the radius and \( h(x) \) is the height of the shell: \( p(x) = x \) and \( h(x) = 3x^2 - x^3 \).

Step 4: Set Up the Integral

Insert the expressions for \( p(x) \) and \( h(x) \) into the shell method integral: \[ V = 2\pi \int_{0}^{3} x(3x^2 - x^3) \, dx \].

Step 5: Simplify the Integrand

Expand the integrand: \[ x(3x^2 - x^3) = 3x^3 - x^4 \]. So the integral becomes \[ V = 2\pi \int_{0}^{3} (3x^3 - x^4) \, dx \].

Step 6: Compute the Integral

Calculate the integral:\[\int_{0}^{3} (3x^3 - x^4) \, dx = \left[ \frac{3x^4}{4} - \frac{x^5}{5} \right]_{0}^{3} \].Evaluate these expressions at the bounds and subtract: \[ \frac{3(3)^4}{4} - \frac{(3)^5}{5} - \left[0\right] = \frac{3(81)}{4} - \frac{243}{5} \].

Step 7: Calculate the Definite Integral Result

Compute the numeric result: \[ \frac{243}{4} - \frac{243}{5} = \frac{1215}{20} - \frac{972}{20} = \frac{243}{20} \].

Step 8: Multiply by Constant Factor

Finally, multiply this result by \( 2\pi \) to find the volume: \[ V = 2\pi \times \frac{243}{20} = \frac{486\pi}{20} = \frac{243\pi}{10} \].

Key Concepts

Shell Method

The shell method is a powerful technique in calculus to find the volume of a solid of revolution. This occurs when a plane region is revolved around an axis, generating a three-dimensional object. In this method, we visualize the solid being composed of many thin cylindrical shells stacked together. Each shell contributes a small volume, and their integration offers the total volume of the shape.
The fundamental formula for the shell method is:

• \[ V = 2\pi \int_{a}^{b} p(x) h(x) \, dx \]

where \( p(x) \) is the distance from the shell to the axis of rotation (the radius), and \( h(x) \) is the height of the shell. Using this method is particularly advantageous when the region is rotated around a vertical line, such as the \( y \)-axis.

Cylindrical Shells

Cylindrical shells play a central role in calculating volumes using the shell method. A cylindrical shell is essentially a hollow cylinder, or tube, with very thin thickness, created when a vertical strip of the area is revolved around an axis.
In our exercise, when the curve \( y = 3x^2 - x^3 \) is revolved around the \( y \)-axis, each small slice (a slice along the \( x \)-axis) contributes a shell. The height of each shell corresponds to the value of \( y \) (i.e., \( h(x) = 3x^2 - x^3 \)), and the radius is simply the \( x \)-coordinate itself, i.e., \( p(x) = x \).

• The surface area of a cylindrical shell is determined by its circumference and height, given by: \( 2\pi \times \text{radius} \times \text{height} \)
  Incorporating thickness, the shell’s volume becomes an integral part of the solid's total volume, calculated through integration over the specified interval.

Definite Integral

The definite integral is a cornerstone of calculus, representing the accumulated quantity of a function over a specific interval. It is essential in computing areas, volumes, and many other geometric and physical properties.
When using the shell method, the definite integral helps us sum the contributions of infinitely many thin cylindrical shells over a given domain from \( a \) to \( b \). For our problem, the interval is from \( x = 0 \) to \( x = 3 \).

• The expression \( \int_{0}^{3} (3x^3 - x^4) \, dx \) calculates the total contribution of the shells between these bounds.
• Applying the Fundamental Theorem of Calculus enables the evaluation of the integral, providing the net volume before multiplying by the constant factor, \( 2\pi \).

Calculus

Calculus is the branch of mathematics focused on change through derivatives and accumulation through integrals. In this exercise, calculus plays a critical role in finding the volume of the region revolved around the \( y \)-axis.
Beyond the specific shell method, calculus encompasses the techniques needed for setting up and evaluating the integrals. From determining the function's roots (where it intersects the \( x \)-axis) to writing the integral for the shell method, calculus offers tools like differentiation and integration to tackle complex problems.

• The differentiation helps identify critical points and analyze the behavior of functions, such as when finding intersections.
• The integration process then calculates total quantities, like the volume, derived from revolving the function’s region about an axis.

Calculus provides a structured approach to solve problems involving continuous change and accumulation, essential for understanding physical and geometric queries effectively.