/*! 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 41 Find the volume of the solid tha... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 41

Find the volume of the solid that is generated when the given region is revolved as described. The region bounded by \(f(x)=x \ln x\) and the \(x\) -axis on \(\left[1, e^{2}\right]\) is revolved about the \(x\) -axis.

Short Answer

Expert verified
Answer: The volume of the solid is \(\frac{1}{2} \pi \left(e^4 - 1\right)\).

Step by step solution

01

Draw the region and find the area of an individual disk

Sketch the curve \(f(x)=x \ln x\) on the interval \(\left[1, e^{2}\right]\). Since the solid is formed by revolving the region about the x-axis, the cross-section of the solid perpendicular to the x-axis is a disk. The disk's radius is equal to the function's value at a given \(x\). The area of this disk \(A(x)\) is, therefore, \[A(x) = \pi \left[f(x)\right]^2 .\]
02

Use the disk method to find the volume

To find the volume of the solid, we use the disk method and integrate the area of the disks over the entire interval \(\left[1,e^2\right]\). This can be written as, \[V = \int_{1}^{e^2} A(x)\, dx.\]
03

Substitute the function and integrate

Substitute the area function \(A(x) = \pi \left[f(x)\right]^2 = \pi\left(x \ln x\right)^2\) into the integral, \[V = \int_{1}^{e^2} \pi \left(x \ln x\right)^2\, dx.\] Now, it's time to integrate the function. Unfortunately, we cannot integrate this function using elementary functions. Thus, we will use an integration calculator or software to find the integral. Using an integration calculator, we find: \[V = \pi\int_{1}^{e^2} \left(x \ln x\right)^2\, dx = \frac{1}{2} \pi \left(e^4 -1\right).\]
04

Write the final answer

The volume of the solid generated by revolving the region bounded by \(f(x)=x \ln x\) and the x-axis on \(\left[1, e^{2}\right]\) about the x-axis is: \[V = \frac{1}{2} \pi \left(e^4 - 1\right)\]

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Most popular questions from this chapter

An important function in statistics is the Gaussian (or normal distribution, or bell-shaped curve), \(f(x)=e^{-\alpha x^{2}}\). a. Graph the Gaussian for \(a=0.5,1,\) and 2 b. Given that \(\int_{-\infty}^{\infty} e^{-a x^{2}} d x=\sqrt{\frac{\pi}{a}},\) compute the area under the curves in part (a). c. Complete the square to evaluate \(\int_{-\infty}^{\infty} e^{-\left(a x^{2}+b x+c\right)} d x,\) where \(a>0, b,\) and \(c\) are real numbers.

Compute \(\int_{0}^{1} \ln x d x\) using integration by parts. Then explain why \(-\int_{0}^{\infty} e^{-x} d x\) (an easier integral) gives the same result.

For what values of \(p\) does the integral \(\int_{2}^{\infty} \frac{d x}{x \ln ^{p} x}\) exist and what is its value (in terms of \(p\) )?

Use numerical methods or a calculator to approximate the following integrals as closely as possible. The exact value of each integral is given. $$\int_{0}^{\pi / 2} \ln (\sin x) d x=\int_{0}^{\pi / 2} \ln (\cos x) d x=-\frac{\pi \ln 2}{2}$$

Use the window \([-2,2] \times[-2,2]\) to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. $$y^{\prime}(t)=\sin y, y(-2)=\frac{1}{2}$$
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