Problem 32 - Calculus Early Transcendentals

Question

Use the shell method to find the volume of the following solids. A hole of radius \(r \leq R\) is drilled symmetrically along the axis of a bullet. The bullet is formed by revolving the parabola \(y=6\left(1-\frac{x^{2}}{R^{2}}\right)\) about the \(y\) -axis, where \(0 \leq x \leq R\).

Step-by-Step Solution

Answer

Using the shell method, we found that the volume of the bullet-shaped solid with a hole drilled symmetrically along the axis is given by: \[V = 6\pi\left(\frac{R^2}{2} + Rr - \frac{R^2}{4} - \frac{rR}{3}\right).\]

Step 1: Find the representative shell

Let's consider a vertical strip at distance \(x\) from the y-axis with thickness \(dx\). The strip, when revolved around the y-axis, will form a cylindrical shell with an inner radius \(r\) and an outer radius \(x\).

Step 2: Determine the height and thickness of the shell

The height of the cylindrical shell is equal to the function \(y = 6\left( 1 - \frac{x^2}{R^2} \right)\). The thickness of the shell is \(dx\), the small change in \(x\).

Step 3: Calculate the volume of a single shell

The volume of a single cylindrical shell is given by the formula: \(V = 2\pi \cdot \text{average radius} \cdot \text{height} \cdot \text{thickness}\). The average radius is \(\frac{x + r}{2}\), so the volume of a single shell is given by \(dV = 2\pi \frac{x + r}{2} \cdot 6\left( 1 - \frac{x^2}{R^2} \right) \cdot dx\).

Step 4: Set up the integral

We want to find the volume of the entire solid, so we integrate the volume of a single shell from \(x = 0\) to \(x = R\): \[\int_{0}^{R} 2\pi \frac{x + r}{2} \cdot 6\left( 1 - \frac{x^2}{R^2} \right) dx.\]

Step 5: Evaluate the integral

Simplify the expression: \[\pi \int_{0}^{R} (x + r) \cdot 6\left( 1 - \frac{x^2}{R^2} \right) dx = 6\pi\int_{0}^{R} (x + r) \left( 1 - \frac{x^2}{R^2} \right) dx.\] Now, compute the integral: \begin{align*} 6\pi\int_{0}^{R} (x + r) \left( 1 - \frac{x^2}{R^2} \right) dx &= 6\pi\int_{0}^{R} (x + r - x\frac{x^2}{R^2} - r\frac{x^2}{R^2}) dx \\ &= 6\pi\int_{0}^{R} (x + r - \frac{x^3}{R^2} - \frac{rx^2}{R^2}) dx \\ &= 6\pi\Bigg[ \frac{x^2}{2} + rx - \frac{x^4}{4R^2} - \frac{rx^3}{3R^2}\Bigg]_{0}^{R} \\ &= 6\pi\left(\frac{R^2}{2} + Rr - \frac{R^4}{4R^2} - \frac{RrR^2}{3R^2}\right) \\ &= 6\pi\left(\frac{R^2}{2} + Rr - \frac{R^2}{4} - \frac{rR}{3}\right). \end{align*}

Step 6: Final answer

The volume of the solid with a hole drilled symmetrically along the axis is given by: \[V = 6\pi\left(\frac{R^2}{2} + Rr - \frac{R^2}{4} - \frac{rR}{3}\right).\]

Key Concepts

Volume of Solids

Volume is an essential concept in geometry that measures the amount of space an object occupies. When it comes to solids, such as cylinders, cones, and spheres, determining the volume can be crucial in various fields like engineering and architecture. Calculating the volume can involve a wide range of methods, depending on the solid's shape:

• Regular Solids: Can often be solved directly using well-known formulas. For example, the volume of a cylinder is given by \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height.
• Irregular Solids: May require decomposition into simpler shapes or using calculus-based methods.

The shell method is one such calculus-based technique used for finding volumes of solids, especially useful when dealing with solids of revolution.

Integral Calculus

Integral calculus is a branch of mathematics concerned with the concept of integration, which is used to compute quantities like areas, volumes, and other quantities dependent on accumulation. There are two main types of integrals:

• Definite Integrals: Used to find the accumulation of quantities, such as the total area under a curve from point \( a \) to point \( b \). An example of a definite integral is \( \int_a^b f(x) \, dx \).
• Indefinite Integrals: Represent the general form of antiderivatives of functions, without specific bounds.

In volume calculations, definite integrals are key. When using the shell method, we set up a definite integral to calculate the total volume by summing up the infinitesimally small shells over a specified range. For instance, in our exercise, we integrate from \( x = 0 \) to \( x = R \) to find the volume of the solid, which involves the integral \[ 6\pi \int_{0}^{R} (x + r - \frac{x^3}{R^2} - \frac{rx^2}{R^2}) \, dx \].

Revolving Solids

Revolving solids is a technique in which a two-dimensional region is revolved around an axis to create a three-dimensional solid. This is a popular method in physics and engineering to analyze objects formed from rotating shapes. The process involves:

• Choosing a region in the plane, typically bounded by curves or lines.
• Revolving this region around a fixed axis, such as the x-axis or y-axis.
• This revolution generates a solid, and various techniques, like the shell method, can be used to find its volume.

For example, in the given exercise, a parabola defined as \( y = 6\left(1\,-\frac{x^{2}}{R^{2}}\right) \) is revolved around the \( y \)-axis. This action results in creating a 3D bullet-shaped object.