Problem 70 - Calculus Early Transcendentals

Question

Suppose \(R\) is the region bounded by \(y=f(x)\) and \(y=g(x)\) on the interval \([a, b],\) where \(f(x) \geq g(x)\). a. Show that if \(R\) is revolved about the vertical line \(x=x_{0}\), where \(x_{0}b ?\)

Step-by-Step Solution

Answer

Answer: The main difference in the formula is related to the distance between each shell's center and the vertical line of rotation. If the line of rotation is to the left of the region, i.e., \(x_0 < a\), the formula for the volume is \(V = \int_{a}^{b} 2 \pi (x-x_0)(f(x)-g(x)) dx\). If the line of rotation is to the right of the region, i.e., \(x_0 > b\), the formula for the volume is \(V = \int_{a}^{b} 2 \pi (x_{0}-x)(f(x)-g(x)) dx\). In both cases, 'R' is the region bounded by \(y=f(x)\) and \(y=g(x)\) on the interval \([a, b]\) and where \(f(x) \geq g(x)\).

Step 1: a. Deriving the volume formula using the shell method

We know that the region \(R\) is bounded by \(f(x)\) and \(g(x)\) on the interval \([a, b]\). We will now calculate the volume of the solid generated by revolving \(R\) about the vertical line \(x = x_0\) where \(x_0 < a\) using the shell method. The shell method works by summing up infintessimal cylinders (shells) formed from revolving each strip of the region R around the vertical line at \(x = x_0\). The height of each shell is the difference of the two functions \(h(x)= f(x)-g(x)\), and the distance between each shell's center and the vertical line is \(s(x) = x - x_0\). The circumference of each shell is \(2\pi s(x)\), and thus the lateral surface area of the shell is \(2\pi s(x)h(x)\). Now we will sum up the volumes of all these infinitesimally thin shells by integrating over the interval \([a, b]\) with respect to x: \(V = \int_{a}^{b} 2 \pi s(x) h(x) dx = \int_{a}^{b} 2 \pi\left(x-x_{0}\right)(f(x)-g(x)) d x\) This shows the volume of the resulting solid when \(R\) is revolved around the line \(x = x_0\) and \(x_0 < a\).

Step 2: b. Modifying the volume formula for a different axis of rotation

Now, we will consider the case when \(x_0 > b\). In this situation, the only thing changing is the way we calculate the distance between the center of each shell and the vertical line at \(x = x_0\). Previously, this distance was simply \(x - x_0\), but now, since \(x_0 > b\), our shells will be formed by revolving from right to left rather than from left to right. This means the distance between the center of each shell and the vertical line is now \(x_0 - x\). So the volume formula gets modified as follows: \(V = \int_{a}^{b} 2 \pi s(x) h(x) dx = \int_{a}^{b} 2 \pi\left(x_{0} - x\right)(f(x)-g(x)) d x\) This new formula calculates the volume of the resulting solid when \(R\) is revolved around the line \(x = x_0\) and \(x_0 > b\).

Key Concepts

Volume of Solids of Revolution

The volume of solids of revolution is a fundamental concept in calculus used to find the volume of a 3D object that is created by rotating a 2D shape around a specific axis. In our given problem, this shape is generated by rotating a bounded region around a vertical line, such as one defined by the Shell Method. This process transforms the region into a three-dimensional object, resembling shapes like vases or barrels.

When using the Shell Method for revolutions around vertical lines, we consider the height and circumference of these cylindrical shells:

• Height of each shell: It is the vertical distance between the functions defining the region, given as \( f(x) - g(x) \) where \( f(x) \) is the function on top.
• Distance from axis of rotation: Typically represented as \( x - x_0 \), this measures how far each shell is from the axis of revolution.
• Volume Calculation: By aggregating these thin shells, we form the full volume through integration over the specified interval.

This method is particularly useful when functions are complicated or when the axis of rotation is not one of the major axes.

Definite Integrals

Definite integrals serve as a powerful tool in calculus to accumulate quantities over an interval. In the context of the Shell Method, they allow us to sum up the volumes of the infinitesimally small cylindrical shells across the interval \[ [a, b] \].

The integral operation assembles these tiny shell volumes to produce the total volume of the solid. The formula is given by:

• \( V = \int_{a}^{b} 2 \pi (x-x_0)(f(x)-g(x)) \, dx \).

This integral indicates:

• The Expression \(2\pi(x-x_0)(f(x)-g(x))\) represents the lateral surface area of each thin shell.
• The Limits \(a\) and \(b\) specify the range over which our region is defined and thus determines the range of shells to sum.

Such integrations are crucial as they provide a way to handle continuous variables and can be calculated using either analytical or numerical methods.

Axis of Rotation

The axis of rotation plays a critical role in determining the formation and dimensions of the solid when considering revolutions in calculus problems. This axis is essentially the line about which the two-dimensional region spins to create a three-dimensional solid.

For the Shell Method discussed in the problem, the axis of rotation is defined by the vertical line \( x = x_0 \).

• If \( x_0 < a \), the shells are formed to the right of the axis, and the distance from the axis is calculated as \( x - x_0 \).
• However, if \( x_0 > b \), the rotation happens from right to left, altering the distance metric to \( x_0 - x \).

Adjusting calculations to suit different axes of rotation is crucial in correctly determining the volume. This requires a careful understanding of how the axis's position relative to the region changes the geometry of the formed cylindrical shells.
By thoroughly understanding these aspects, you can precisely apply integration techniques to calculate the volumes of solids of revolution in various configurations.