/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 The solid formed by revolving th... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 39

The solid formed by revolving the region bounded by the graphs of \(y=x, y=4,\) and \(x=0\) about the \(x\) -axis

Short Answer

Expert verified
\[\frac{64}{3}\pi\]

Step by step solution

01

Graph the functions

To better understand the region, plot the functions \(y=x\), \(y=4\), and \(x=0\) on a coordinate plane. You will see that this creates a triangle with vertices at (0,0), (0,4) and (4,4).
02

Set up the integral

We are rotating around the x-axis, so we will use the formula for the volume of a solid of revolution: \[V = \pi\int_{a}^{b}(f(x))^2 dx\] where \((f(x))^2\) is the radius of the disk at \(x\). This is the difference between our functions. Since the integral goes from \(x=a\) to \(x=b\), we can see from our graph that \(a=0\) and \(b=4\). Our radius function \(f(x)\) here should be \(4-x\) as the upper function is the constant \(y=4\), while the lower function is \(y=x\). This gives us \[V=\pi\int_{0}^{4} ((4-x))^2 dx\].
03

Evaluate the integral

The integral \(\int_{0}^{4} ((4-x))^2 dx\) is straightforward to solve. Expand the square to get \(\int_{0}^{4} (16 - 8x + x^2) dx\). Split this into three separate integrals and solve it. The answer will be \[\frac{64}{3}\pi\].

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that deals with rates of change (differential calculus) and accumulation of quantities (integral calculus). Integral calculus, in particular, is powerful for calculating areas, volumes, and lengths that are otherwise difficult to compute. For example, it can be used to find the volume of a solid of revolution - a solid created by rotating a two-dimensional shape around an axis.

The exercise involves using calculus to compute such a volume, wherein the initial step is to graph the functions to identify the region of interest. After graphing the regions, calculus is used to set up an integral that will represent the volume of the 3D shape created by rotating the region around the x-axis. The comprehension of these graphs and function behaviors is key for setting up the appropriate definite integral to calculate the volume.
Definite Integration
Definite integration is a concept within calculus that allows you to find the accumulated quantity over a particular interval. The process involves finding the integral of a function between two points, which gives the net area under the curve of the function within those points. In the provided exercise, definite integration comes into play when determining the volume of the solid of revolution.

The definite integral \[V = \pi\int_{a}^{b}(f(x))^2 dx\] requires proper understanding of the limits of integration, which are the values of 'a' and 'b'. In this problem, the area that rotates around the x-axis is a triangle, bounded by 'a=0' and 'b=4'. By setting up the definite integral within the right boundaries, and integrating over this interval, students determine the exact volume of the resulting solid.
Disk Method
The disk method is a technique in calculus for finding the volume of a solid of revolution when a region in a plane is revolved around an axis. It involves slicing the solid into thin disks perpendicular to the axis of rotation and then summing the volumes of these disks. Each disk's volume is that of a cylinder, \[\text{Volume of a disk} = \pi \times (\text{radius})^2 \times (\text{thickness})\].

In practice, the 'thickness' is represented by an infinitesimally small change in 'x' or 'dx' when revolving around the x-axis. We calculate the accumulation of these infinitesimal volumes using a definite integral. Regarding the exercise, the radius of each disk in the solid is determined by the function \[f(x) = (4-x)\], which changes as 'x' changes from 0 to 4. The method simplifies the complex problem of finding the volume of a 3D solid into a manageable one-dimensional integral that calculates the sum of infinitely many thin disks' volumes.

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Most popular questions from this chapter

Hydraulic Press In Exercises \(45-48,\) use the integration capabilities of a graphing utility to approximate the work done by a press in a manufacturing process. A model for the variable force \(F\) (in pounds) and the distance \(x\) (in feet) the press moves is given. \(F(x)=1000[1.8-\ln (x+1)] \quad 0 \leq x \leq 5\)

$$\pi \int_{0}^{2}\left[16-(2 y)^{2}\right] d y=2 \pi \int_{0}^{4} x\left(\frac{x}{2}\right) d x$$

Finding Arc Length In Exercises \(21-30\) , (a) sketch the graph of the function, highlighting the part indicated by the given interval, (b) write a definite integral that represents the 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 arc length.arc length of the curve over the indicated interval and observe $$y=\ln x, \quad[1,5]$$

Finding Arc Length In Exercises \(21-30\) , (a) sketch the graph of the function, highlighting the part indicated by the given interval, (b) write a definite integral that represents the 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 arc length.arc length of the curve over the indicated interval and observe $$y=\frac{1}{x+1}, \quad[0,1]$$

Think About It Match each integral with the solid whose volume it represents and give the dimensions of each solid. (a) Right circular cone (b) Torus (c) Sphere (d) Right circular cylinder (e) Ellipsoid $$\begin{array}{l}{\text { (i) } 2 \pi \int_{0}^{r} h x d x} \\ {\text { (ii) } 2 \pi \int_{0}^{t} h x\left(1-\frac{x}{r}\right) d x}\end{array}$$ $$\begin{array}{l}{\text { (iii) } 2 \pi \int_{0}^{r} 2 x \sqrt{r^{2}-x^{2}} d x} \\ {\text { (iv) } 2 \pi \int_{0}^{b} 2 a x \sqrt{1-\frac{x^{2}}{b^{2}}} d x} \\ {\text { (v) } 2 \pi \int_{-r}^{r}(R-x)\left(2\sqrt{r^{2}-x^{2}}\right) d x}\end{array}$$
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