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Finding the Volume of a Solid In Exercises \(23-26,\) use the shell method to find the volume of the solid generated by revolving the plane region about the given line. $$ y=\frac{1}{3} x^{3}, \quad y=6 x-x^{2}, \quad \text { about the line } x=3 $$

Short Answer

Expert verified
The volume of the solid generated by revolving the area bounded by the curves \(y=\frac{1}{3} x^{3}\) and \(y=6 x-x^{2}\) about line \(x=3\) calculated using the shell method is obtained by evaluating the set up integral \(V = 2\pi \int_0^3 (3-x) (6x-x^2 -\frac{1}{3} x^{3}) dx + 2\pi \int_3^6 (x-3) (6x-x^2 -\frac{1}{3} x^{3}) dx\).

Step by step solution

01

Sketch the Area

Start by sketching the area enclosed by the given functions \(y=\frac{1}{3} x^{3}\) and \(y=6 x-x^{2}\). Also, make a note of the line \(x=3\) around which the curves will be revolved.
02

Calculate Intersection points

Determine the points of intersection between these two curves. These points are found by equating the two functions and solving for \(x\):\n\(\frac{1}{3} x^{3}=6 x-x^{2}\) simplifies to \(x^{3}-18x + 9x^{2}=0\), which gives \(x=0, x=3, x=6\). These are the limits of integration.
03

Setup integral using Shell Method

Next we can setup the integral using the Shell Method. The formula for the volume \(V\) of a solid of revolution using the shell method is given by \(V = 2\pi \int_a^b (radius)(height) dx\), where [a,b] are the limits of the integration. The radius is the distance from the shell to the axis of rotation, and the height is the difference between the top and bottom curve. Here, the radius is \(|x-3|\) and the height is \(|(6x-x^2) - (x^3/3)|\). So, the integral becomes: \[V = 2\pi \int_0^6 |x-3| |(6x-x^2) -\frac{1}{3} x^{3}| dx\].
04

Evaluate the Integral

Solving this integral will give the volume of the solid. Due to the absolute values, the integral needs to be split into three parts: from 0 to 3, from 3 to 6, bits. It will become: \(V = 2\pi \int_0^3 (3-x) (6x-x^2 -\frac{1}{3} x^{3}) dx + 2\pi \int_3^6 (x-3) (6x-x^2 -\frac{1}{3} x^{3}) dx\). Evaluating this integral will give the requested result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solid of Revolution
A solid of revolution is a three-dimensional shape created by rotating a two-dimensional area around an axis. This process transforms a flat shape into a voluminous object, much like spinning clay on a potter's wheel forms a vase. In calculus, we often create these solids by revolving a region outlined by curves on a graph around a specified line, referred to as the axis of rotation. The result is a symmetric shape, such as a cylinder, cone, or more complex forms depending on the curves involved.
For instance, take a rectangle on the coordinate plane; if spun around one of its sides, it forms a cylindrical shape. Similarly, revolving complex regions defined by functions can create an array of intricate solids. Recognizing how to visualize these rotations is critical for understanding the calculations that follow when determining the solid's volume.
Volume of Solid Calculus
Calculating the volume of solids in calculus can be done through several methods, including the shell method and the disk/washer method. These techniques involve setting up and evaluating integrals to find the volume of irregularly shaped objects that do not have simple geometric formulas. These calculus-based methods provide a way to partition the solid into infinitesimally thin pieces and then sum the volume of these pieces to find the total volume.
When employing the shell method as seen in the exercise, cylindrical shells are envisioned to make up the solid's volume. We consider the volume of each thin shell and integrate to find the cumulative volume. It's analogous to summing up the volumes of numerous hollow tubes of different radii and heights to get the total volume. Calculus gives us the tools to perform this summation across a continuous range.
Calculating Volume with Integrals
Integrals are at the heart of computing volumes in calculus. They enable us to add up an infinite number of infinitesimally thin slices or shells to find the total volume of a solid. The integration process effectively calculates the limit of increasingly finer sums, akin to adding up the volume of infinitely thin layers that comprise the solid.
For the shell method, we set up an integral where the integrand represents the volume of an individual shell, multiplying the shell's circumference (for lateral surface area) by its height and by the thickness of the shell, which in the limit is an infinitesimal width represented by dx or dy, depending on the axis of rotation. By integrating this expression over the specified interval, we accumulate the volume of all these shells to get the total volume. Thus, integrals serve as a key tool in transitioning from a sum of parts to the whole.
Limits of Integration
The limits of integration are the boundaries within which we integrate to determine the quantity of interest—in this case, the volume of a solid. They are derived from the intersecting points of the curves or from the endpoints of the region being rotated. In our exercise, we find the limits of integration by solving for the points where the two functions intersect, because these points mark the start and end of the solid along the axis of revolution.
Specific to the shell method, these limits help us establish where the cylindrical shells begin and end along the axis. When using the given curves and the axis of rotation, the volume integral has to be calculated between these critical points to capture the complete solid. Understanding how to correctly determine these limits ensures that the integration process accounts for the entire solid, allowing for an accurate calculation of volume.

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Most popular questions from this chapter

Volume of an Ellipsoid Consider the plane region bounded by the graph of $$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$$ where \(a>0\) and \(b>0 .\) Show that the volume of the ellipsoid formed when this region is revolved about the \(y\) -axis is $$\frac{4}{3} \pi a^{2} b$$ What is the volume when the region is revolved about the \(x\) -axis?

Lifting a Chain In Exercises \(25-28\) , consider a 20 -foot chain that weighs 3 pounds per foot hanging from a winch 20 feet above ground level. Find the work done by the winch in winding up the specified amount of chain. Wind up the entire chain.

Finding Arc Length In Exercises \(17-26,\) (a) sketch the graph of the function, highlighting the part indicated by the given interval, (b) find a definite integral that represents the arc length of the curve over the indicated interval and observe that the integral cannot be evaluated with the techniques studied so far, and (c) use the integration capabilities of a graphing utility to approximate the are length. $$ y=\ln x, \quad 1 \leq x \leq 5 $$

Boyle's Law In Exercises 37 and 38 , find the work done by the gas for the given volume and pressure. Assume that the pressure is inversely proportional to the volume. (See Example \(6 . )\) A quantity of gas with an initial volume of 2 cubic feet and a pressure of 1000 pounds per square foot expands to a volume of 3 cubic feet.

Think About It Consider the equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\) (a) Use a graphing utility to graph the equation. (b) Set up the definite integral for finding the first-quadrant arc length of the graph in part (a). (c) Compare the interval of integration in part (b) and the domain of the integrand. Is it possible to evaluate the definite integral? Is it possible to use Simpson's Rule to evaluate the definite integral? Explain. (You will learn how to evaluate this type of integral in Section \(8.8 . )\)

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