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Chapter 6: Problem 53

The region bounded by the graphs of \(y=\ln x, y=0\), \(x=1\), and \(x=e\) is revolved about the \(x\) -axis. Find the volume of the solid generated.

Short Answer

Expert verified
The volume of the solid generated by revolving the region bounded by the graphs of \(y=\ln x, y=0, x=1\), and \(x=e\) about the x-axis is \(V = \pi(2 - e)\).

Step by step solution

01

Find the radius of the disk

In this problem, the radius of the disk is given by the distance between the x-axis and the curve \(y=\ln x\). So the radius r(x) is exactly the value of y-coordinate of the curve, r(x) = \(\ln x\).
02

Find the cross-sectional area of the disk

We can find the cross-sectional area of an infinitesimally thin disk in the region by using the formula for the area of a circle, which is \(A = \pi r^2\). In our case, the area of the disk is given by: \[A(x) = \pi (\ln x)^2\]
03

Set up the integral to find the volume

To find the volume of the solid generated, we need to integrate the cross-sectional area from the leftmost to the rightmost bounds of the region along the x-axis, which are x = 1 and x = e. So, the integral to find the volume is: \[V = \int_1^e \pi (\ln x)^2 dx\]
04

Evaluate the integral

To evaluate this integral, we can use integration by substitution or u-substitution. Let's set: \(u = \ln x => x = e^u => dx = e^u du\) Now, we need to change the limits of integration. When x = 1: \[u = \ln (1) = 0\] When x = e: \[u = \ln (e) = 1\] Now, let's substitute u into our integral and solve for V: \[V = \int_0^1 \pi u^2 e^u du\]
05

Integration by parts

Let's further break down the integral using integration by parts: Let: \(v = u^2 => dv = 2u du\) \(dw = e^u du => w = e^u\) Now, applying the integration by parts formula \(\int udw = uw - \int wdv\), we get: \[V = \pi\left[ u^2e^u\Big|_0^1 - 2 \int_0^1 ue^u du\right]\] Now, let's apply integration by parts again: Let: \(v = u => dv = du\) \(dw = e^u du => w = e^u\) We get: \[V = \pi\left[ u^2e^u\Big|_0^1 - 2\left(ue^u\Big|_0^1 - \int_0^1 e^u du\right) \right]\]
06

Evaluate the remaining integral

Now, we just have to evaluate the remaining integral, \(\int_0^1 e^u du\), which gives: \[\int_0^1 e^u du = e^u\Big|_0^1 = e - 1\]
07

Combine all terms and calculate the volume

Now, substituting the integral we just solved back into the expression for V, we get: \[V = \pi\left[ u^2e^u\Big|_0^1 - 2\left(ue^u\Big|_0^1 - (e - 1) \right) \right]\] Evaluating the terms inside the brackets and simplifying: \[V = \pi\left[ (1 \cdot e) - 2\left((1 \cdot e) - (e - 1) \right) \right]\] \[V = \pi\left[ e - 2e + 2(e - 1) \right]\] \[V = \pi(2 - e)\] Therefore, the volume of the solid generated by revolving the region bounded by the graphs of \(y=\ln x, y=0, x=1\), and \(x=e\) about the x-axis is \(V = \pi(2 - e)\).

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