Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

FInd the volume of the solid generated when the region \(R\) bounded by the given curves is revolved about the indicated axis. Do this by performing the following steps. (a) Sketch the region \(R\). (b) Show a typical rectangular slice properly labeled. (c) Write a formula for the approximate volume of the shell generated by this slice. (d) Set up the corresponding integral. (e) Evaluate this integral. \(y=\frac{1}{4} x^{3}+1, y=1-x, x=1 ;\) about the \(y\) -axis

Step-by-Step Solution

Answer

The volume is \( \frac{23\pi}{30} \).

Step 1: Sketch the Region R

Start by sketching the curves given in the problem. Plot the functions \( y = \frac{1}{4}x^3 + 1 \) and \( y = 1 - x \) on the xy-plane. Identify the intersection points of these curves, particularly focusing on the relevant domain of \( x \) from 0 to 1, since this contains the bounded region of interest. The line \( x = 1 \) acts as a boundary as well.

Step 2: Typical Rectangular Slice

In order to apply the shell method, we consider a vertical slice at a point \( x \). The height of this slice is the vertical distance between the curves: top function minus bottom function, i.e., \( (1 - x) - (\frac{1}{4}x^3 + 1) = -x - \frac{1}{4}x^3 \). The thickness of this slice is \( \Delta x \).

Step 3: Formula for Volume of Shell

When this slice is revolved around the \( y \)-axis, it forms a cylindrical shell. The radius of the shell is \( x \), and its height is \( -x - \frac{1}{4}x^3 \). The approximate volume \( \Delta V \) of the shell is calculated using:\[\Delta V = 2\pi \, \text{(radius)} \, \text{(height)} \, \Delta x = 2\pi x(-x - \frac{1}{4}x^3)\Delta x.\]

Step 4: Set Up the Integral

To find the exact volume, integrate the formula for \( \Delta V \) with respect to \( x \), evaluating from 0 to 1. This gives:\[V = \int_0^1 2\pi x(-x - \frac{1}{4}x^3)\, dx.\]

Step 5: Evaluate the Integral

Expand the integrand and simplify it:\[V = 2\pi \int_0^1 (-x^2 - \frac{1}{4}x^4)\, dx = 2\pi \left( -\int_0^1 x^2 \, dx - \frac{1}{4} \int_0^1 x^4 \, dx \right).\]Calculate each part of the integral:\[\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3},\]\[\int_0^1 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5}.\]Substitute back:\[V = 2\pi \left( -\frac{1}{3} - \frac{1}{4} \times \frac{1}{5} \right) = 2\pi \left( -\frac{1}{3} - \frac{1}{20} \right) = 2\pi \left( -\frac{20 + 3}{60} \right) = -2\pi \times \frac{23}{60}.\]Therefore, the volume of the solid is:\[V = -\frac{46\pi}{60} = -\frac{23\pi}{30}.\]Since volume cannot be negative, take the positive value.

Key Concepts

Shell Method

The Shell Method is a way to find the volume of a solid of revolution. It is especially useful for solids generated by revolving a region around a line parallel to the axis of logs. The method involves using cylindrical shells or layers, which are essentially thin hollow tubes. Imagine peeling an onion where each layer represents a shell. Each shell is formed by revolving a tiny slice of the region perpendicular to the axis of revolution.

When using the shell method, follow these steps:

• Identify the axis of rotation. If it's along the y-axis, vertical slices are used.
• Consider a rectangular slice of the area you want to rotate. This slice will create a cylindrical shell.
• Determine the radius of the shell which is the distance from the axis of rotation to the slice. If revolving around the y-axis, the radius is generally the x-coordinate of the slice.
• Calculate the height of the shell by finding the difference between the functions defining the top and bottom of the region.
• The formula for the shell's volume is: \[ \Delta V = 2\pi \cdot \text{(radius)} \cdot \text{(height)} \cdot \Delta x \]
• Then integrate this expression over the interval that encloses the bounded region to find the total volume.

Integral Calculus

Integral Calculus is a fundamental branch of calculus focusing on the concept of integration. This mathematical operation is essentially the reverse of differentiation. It helps in finding quantities like area, volume, and other accumulative values. Integration allows us to evaluate quantities within a specified range, thereby enabling the determination of definite integrals among other concepts.

The two main types of integrals are definite and indefinite:

• Definite Integrals: These calculate the accumulation from one point to another within a closed interval. They are denoted by integral symbols with upper and lower limits, indicating the interval of integration.
• Indefinite Integrals: These refer to a general equation of an antiderivative without specified limits. They represent a family of functions and include a constant of integration.

In calculus challenges such as volume of solids of revolution, definite integrals are primarily used to evaluate exact volumes. Integral Calculus, thus, provides the mathematical tools to tackle these problems effectively.

Cylindrical Shells

Cylindrical shells are a key component when employing the shell method. In essence, they are the results of revolving a slice of a region about a vertical or horizontal axis. Just as cylinders are the 3D counterparts of circles, cylindrical shells are formed by thin slices of the region which are then rotated.

Important aspects of cylindrical shells include:

• Radius: The radius of a shell is determined by how far the slice is from the axis of rotation. When revolving around the y-axis, it is often the x-value of the slice.
• Height: This is given by the difference between the output values of the top and bottom functions defining the region.
• Thickness: The thickness of each shell is a small change in the x-value or y-value, represented by \( \Delta x \) or \( \Delta y \).

Cylindrical shells provide an elegant way to approximate the volume of complex shapes by integrating many such shells across the domain. Hence, they are central to solving many real-world problems involving revolution-generated solids.

Definite Integrals

Definite Integrals are a core concept in calculus used to calculate the total accumulation of quantities within a specific interval. Unlike indefinite integrals, they provide a single number as the result, representing the computed total from the lower to upper bound of integration.

For evaluating volumes of solids of revolution, definite integrals are essential. They combine the incremental volumes of cylindrical shells to find the entire volume.

• They are written in the form: \[ \int_{a}^{b} f(x) \, dx \]where \( a \) and \( b \) are the limits of integration.
• Definite integrals involve calculating the antiderivative of the function within the given limits, followed by taking the difference between the upper and lower bounds.
• In volume calculations, the function \( f(x) \) often represents some part of the formula for the shell's volume.

Using definite integrals allows for a rigorous method to calculate areas and volumes, solidifying their status as one of the most powerful tools in mathematics.