Problem 35 - Calculus of a Single Variable

Question

Finding the Volume of a Solid In Exercises \(33-36\) (a) use a graphing utility to graph the plane region bounded by the graphs of the equations, and (b) use the integration capabilities of the graphing utility to approximate the volume of the solid generated by revolving the region about the \(y\) -axis. $$ y=\sqrt[3]{(x-2)^{2}(x-6)^{2}}, \quad y=0, \quad x=2, \quad x=6 $$

Step-by-Step Solution

Answer

The volume of the solid can be found by evaluating \( V = \pi \int_{2}^{6} [\sqrt[3]{(x-2)^{2}(x-6)^{2}}]^{2}dx \) using the integration capabilities of the graphing utility.

Step 1: Graphing the Function

Load the function \( y=\sqrt[3]{(x-2)^{2}(x-6)^{2}} \) onto a graphing utility, and plot it between \( x=2 \) and \( x=6 \). The function should be continuous and smooth within this range.

Step 2: Setting Up the Volume Integral

Next, define the formula to approximate the volume of the solid of revolution. Using the disc method, the formula is given by \( V = \pi \int_{a}^{b} [f(x)]^{2}dx \) where \( a \) and \( b \) are the limits of revolution, \(2\) and \(6\) respectively. Here, \( f(x) = \sqrt[3]{(x-2)^{2}(x-6)^{2}} \). So the integral setup will be \( V = \pi \int_{2}^{6} [\sqrt[3]{(x-2)^{2}(x-6)^{2}}]^{2}dx \)

Step 3: Evaluating the Integral

Use the built-in integration capabilities of the graphing utility to evaluate the integral defined in Step 2. This will give you the approximate volume of the solid of revolution.

Key Concepts

Graphing Functions

Graphing functions is a foundational tool that helps us visualize the relationship between variables in a mathematical expression. By plotting a function on a coordinate plane, we can gain insights into its behavior, such as interval of increase or decrease, maximum or minimum points, and symmetry. In our exercise, we are interested in graphing the function \( y=\root{3}\root{(x-2)^{2}(x-6)^2} \).

To graph this function correctly, we input it into a graphing utility, focusing on the interval between \( x=2 \) and \( x=6 \), since we're only interested in these values for our volume calculation. The graph should reveal a region bounded by the curve, the x-axis (\( y=0 \)), and the vertical lines \( x=2 \) and \( x=6 \). This visual representation is crucial because it helps us understand the shape and size of the plane region that will be rotated around the y-axis to form a solid.

Making sure we understand the graph correctly is important because any mistake here would directly affect the accuracy of our volume calculation.

Disc Method Integration

The disc method is an integral calculus technique used to calculate the volume of a solid of revolution. When a region in the plane is rotated around an axis, it creates a three-dimensional object. The disc method slices this object into thin discs perpendicular to the axis of rotation, much like slicing a loaf of bread.

The volume of each disc is approximated by the volume of a cylinder, \( V_{\text{disc}} = \text{A}\times\text{thickness} \), where A is the area of the disc's face, and 'thickness' is an infinitesimally small width of the disc. We're essentially stacking these discs from start to finish of the area's interval to find the whole volume:
\[ V = \text{π} \times \text{thickness} \times \text{\text{Σ}} \text{A} \] For continuous functions and using calculus integration, this becomes:
\[ V = \text{π} \times \text{thickness} \times \text{\text{∫}} \text{A} \] In our exercise, the thickness of each disc is given by a small change in x (dx), and the area A is given by \( \text{π}[f(x)]^2 \). So, the integral for the volume is \( V = \text{π} \text{∫}_{a}^{b} [f(x)]^2dx \), where \( a \) and \( b \) define the bounds of the region being rotated.

Solid Generated by Revolution

A solid of revolution is a three-dimensional object created by rotating a two-dimensional plane area around an axis. The shape of the solid depends on the curve of the function and the axis around which it's rotated. In our case, the plane region bounded by \( y = \root{3}\root{(x-2)^{2}(x-6)^2} \), the x-axis, and the lines \( x=2 \) and \( x=6 \) is being revolved around the y-axis.

It's essential to understand that each point on the function curve traces out a circle as we rotate it around the y-axis. The radius of these circles is determined by the horizontal distance from the y-axis to the curve, which in this instance is the value of our function \( f(x) \) at any given point \( x \). As we integrate along the bounds of \( x \), from 2 to 6, we're effectively summing up the volumes of an infinite number of infinitesimally thin discs to get the volume of the solid.

Using the graphing utility's integration feature, we approximate this sum to find the volume of the solid generated by revolution. This process is not just a powerful application of integral calculus but also a demonstration of how multidimensional shapes can be derived from simple two-dimensional graphs.