Problem 7 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. $$y=\left(1+x^{2}\right)^{-1}, y=0, x=0, \text { and } x=2$$

Step-by-Step Solution

Answer

Question: Find the volume of the solid generated when the region R, bounded by the curves \(y = (1+x^2)^{-1}, y = 0, x = 0\), and \(x = 2\), is revolved around the y-axis using the shell method. Answer: The volume of the solid generated when region R is revolved around the y-axis is \(\pi \ln 5\).

Step 1: Sketch the region

First, we need to sketch the region bounded by the given curves. We can see that the curve \(y = \left(1+x^2\right)^{-1}\) is asymptotic to the x-axis and y-axis. Based on the given constraints, the entire region of interest lies in the first quadrant.

Step 2: Set up the shell

Using the shell method, the volume of the solid, V, can be found using the formula: $$V = 2\pi \int_a^b p(x)h(x)dx$$ where \(p(x) = x\) is the distance from the axis of revolution (the \(y\)-axis in this case) to the strip, and \(h(x) = y\), is the height of the shell, and the integral is computed along the x-axis from \(a\) to \(b\). In our case, \(a = 0\) and \(b = 2\). From the given equation, we have \(h(x) =\left(1 + x^2\right)^{-1}\).

Step 3: Writing the integral

Now we can set up the integral for the volume of revolution around the y-axis: $$V = 2\pi \int_0^2 x\left(1+x^2\right)^{-1} dx$$

Step 4: Evaluating the integral

Now we need to evaluate the integral to find the volume. Using substitution, let: $$u = 1 + x^2 \Rightarrow du = 2x dx$$ We also need to change the limits of integration according to the new variable \(u\): $$x = 0 \Rightarrow u = 1$$ $$x = 2 \Rightarrow u = 5$$ So, our integral becomes $$V = 2\pi \int_1^5 \frac{1}{2} u^{-1}du$$ Now we can integrate: $$V = \pi \int_1^5 u^{-1} du = \pi [\ln |u|]_1^5 = \pi (\ln 5 - \ln 1)$$ Since \(\ln 1 = 0\), the final result is $$V = \pi \ln 5$$ Thus, the volume of the solid generated when the region \(R\) is revolved around the \(y\)-axis is \(\pi \ln 5\).

Key Concepts

Volume of Revolution

When we talk about the "Volume of Revolution," we're describing the volume of a three-dimensional object formed by rotating a two-dimensional region around an axis. One way to find such a volume is the "Shell Method" in calculus.
Rotation creates a solid object from a region. This is similar to how a potter shapes clay on a pottery wheel. Here, the given region is between several curves and must be properly understood for accurate volume measurement.

• The first step in the Shell Method involves visualizing the region. This is essential for setting correct bounds.
• Consider the distance each small "shell" or cylindrical slice travels during rotation. This arc-length is crucial for calculating the entire volume accurately.

Understanding these components provides a framework to apply the Shell Method, and helps us calculate the revolved volume around an axis methodically.

Integration

Integration is the mathematical technique used to find the total accumulative result of a function. When applied in calculus, it's like adding up an infinite number of tiny quantities to find a whole. For finding the volume of a revolution, integration takes the small shells generated by rotation and sums them.
In the Shell Method, integration is especially vital. The formula uses the definite integral to sum slices of a region:
\[ V = 2\pi \int_a^b p(x)h(x) \, dx \]Where:

• \(p(x)\) represents the distance from each shell to the axis of rotation.
• \(h(x)\) denotes the shell's height.
• The limits \(a\) and \(b\) define the interval over which we're summing.
Proper understanding and setup of the integral are crucial for accurate volume calculation in the Shell Method, as each step reflects a critical part of the region being analyzed.

Calculus

Calculus serves as the mathematical language that describes dynamic systems like those found in physical contexts—such as volumes formed by revolution, like in our problem.
Two main branches of calculus are differential and integral calculus. The latter is used here, where its tools allow finding volumes of complex shapes that manual measurement can't easily encompass.
In our problem:

• The curve \(y = (1 + x^2)^{-1}\) is a mathematical description modeled by calculus tools.
• The integration limits from \(0\) to \(2\) provide the region's boundary—understanding these help identify where the substantial part of the function lies.

Thus, calculus provides the toolkit that makes understanding and solving such geometric transformations possible. It builds a robust foundation for understanding the behavior of curves, areas, and, consequently, volumes.