Problem 10 - Calculus of a Single Variable

Question

Finding the Volume of a Solid In Exercises \(1-14,\) use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the \(y\) -axis. Graph cannot copy $$ y=x^{3 / 2}, \quad y=8, \quad x=0 $$

Step-by-Step Solution

Answer

The volume of the solid of revolution is \(\frac{128\pi}{7}\) cubic units.

Step 1: Setup

First, we identify the radii, height and thickness of the cylindrical shells. Here, the thickness of the shells is \(dx\), the height is \(y = x^{3/2}\), and the radius is \(x\). The volume of each shell is \(2\pi\)x(the height )(the thickness), which is \(2\pi\)x(\(x*x^{3/2}*dx\)). We need to sum up these volumes from \(x = 0\) to \(x = (8)^{2/3}\).

Step 2: Formulating the Integral Expression for Volume

The integral for the volume using the shell method, from \(x = 0\) to \(x = 8^{2/3}\) would be: \[V = \int_{0}^{8^{2/3}} 2\pi x x^{3/2} dx\]This gives the sum of these infinitesimally small volumes which gives the total volume of the solid.

Step 3: Evaluating the Integral

Solving the integral,\[V = 2\pi \int_{0}^{8^{2/3}} x^{5/2} dx,\]we use the power rule for integration which states that \(\int x^n dx = (x^{n+1}) / (n+1) + C\). Applying this gives \[V = 2\pi [(2/7)x^{7/2}]_{0}^{8^{2/3}},\]which simplifies to \[V = 2\pi * (2/7)*[2^4 - 0] = 2*64\pi/7 = 128\pi/7.\]

Key Concepts

Integration

Integration is a fundamental concept in calculus, widely used to find areas under curves and accumulate quantities. It is essentially the reverse process of differentiation. In integration, you sum up infinitely small parts to find the whole.

When you perform integration, you work to find the integral of a function. This integral represents all the accumulated quantities as the variable changes. In the context of finding volumes, as in our original exercise, integration helps calculate the sum of numerous infinitesimally small volumes to form the entire solid.

For practical applications, integration is often used in problems involving:

• Area under curves
• Volume of solids
• Center of mass
• Total accumulated change

Cylindrical Shell

The cylindrical shell method is a technique used in calculus to find the volume of a solid of revolution. This method is especially useful when revolving around the vertical axis.

In the shell method, a solid is divided into several hollow cylinders (or shells). Each cylinder has:

• A small thickness, usually denoted as \(dx\) or \(dy\)
• A height, which is the function being rotated
• A radius, which is the distance from the axis of rotation to the shell

The formula to find the volume of one of these shells is:\[\text{Volume} = 2\pi (\text{radius})(\text{height})(\text{thickness})\]Each shell's contribution to the volume is computed using this formula, and integration is used to total the volumes of all such shells across the desired interval.

Power Rule for Integration

The power rule for integration is a simple and essential tool in calculus. It helps us integrate functions of the form \(x^n\). When you integrate such functions, the power rule formula is:
\[\int x^n dx = \frac{x^{n+1}}{n+1} + C\]where \(C\) is the constant of integration.

Let's say you have a function like \(x^{5/2}\). Using the power rule:

• You increase the power by 1, turning it into \(x^{7/2}\).
• Divide by the new power, resulting in \(\frac{x^{7/2}}{7/2}\).

This rule greatly simplifies the process of finding integrals, by providing a straightforward way to handle polynomials.

Volume of Solids of Revolution

Finding the volume of solids of revolution is a common task in calculus. These solids are formed when a 2D region is revolved around an axis.

There are several methods to find these volumes, with the shell method being one of them. This method is beneficial when revolving around the vertical axis, like the \(y\)-axis in the original exercise.

The steps involved in using the shell method include:

• Determining the height and radius of each shell from the function and axis of rotation.
• Calculating the differential volume of a single shell using the formula \(2\pi (\text{radius})(\text{height})(\text{thickness})\).
• Setting up an integral to sum up all those differential volumes over the interval.

This integration gives the total volume of the solid as formed by stacking these shells.