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Chapter 7: Problem 42

Find the volume of the solid that is generated when the given region is revolved as described. The region bounded by \(f(x)=e^{-x}\) and the \(x\) -axis on \([0, \ln 2]\) is revolved about the line \(x=\ln 2\).

Short Answer

Expert verified
Answer: The volume of the solid is \(\frac{3\pi}{8}\).

Step by step solution

01

Set up the integral for the disc method

Since we are revolving about the line \(x=\ln 2\) and the discs are perpendicular to this line, we will consider discs of thickness \(dx\). The volume of a single disc is given by the area of the circle times its thickness, so the volume of each disc is: $$ \pi[f(x)]^2 dx $$ Now, we want to sum the volumes of all the discs from \(x=0\) to \(x=\ln 2\): $$ V = \int_{0}^{\ln 2} \pi[f(x)]^2 dx $$
02

Replace \(f(x)\) with the curve equation

Now we should replace \(f(x)\) with the given equation \(e^{-x}\): $$ V = \int_{0}^{\ln 2} \pi[e^{-x}]^2 dx $$
03

Simplify the integral

Simplify the integral by squaring \(e^{-x}\): $$ V = \int_{0}^{\ln 2} \pi e^{-2x} dx $$
04

Integrate with respect to x

Integrate the integral with respect to x: $$ V = \pi\left[-\frac{1}{2}e^{-2x}\right]_{0}^{\ln 2} $$
05

Evaluate the integral at the limits

Now we need to evaluate the expression at the limits \(0\) and \(\ln 2\): $$ V = \pi\left[-\frac{1}{2}e^{-2(\ln 2)} -\left(-\frac{1}{2}e^{-2(0)}\right)\right] $$
06

Simplify and solve for the volume

Simplify and solve the expression to find the volume of the solid: $$ V = \pi \left[-\frac{1}{2}\left(\frac{1}{2^2}\right)+\frac{1}{2}\right] $$ $$ V=\pi \left[-\frac{1}{8}+\frac{1}{2}\right]=\pi\left(\frac{3}{8}\right) $$ So the volume of the solid is: $$ V=\frac{3\pi}{8} $$

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