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Chapter 6: Problem 92

Find the volume of the solid generated by revolving the unbounded region lying between \(y=-\ln x\) and the \(y\) -axis \((y \geq 0)\) about the \(x\) -axis.

Short Answer

Expert verified
The volume of the solid generated by revolving the unbounded region between \(y=-\ln x\) and the \(y\)-axis (y is greater than or equal to 0) about the \(x\)-axis is 0 cubic units.

Step by step solution

01

Define the Integral

The volume of the solid of revolution generated by revolving the function \(y = -\ln x\) about the \(x\) axis can be calculated using the formula for rotating about the \(x\) -axis: \(V = \pi \int_1^{+\infty} (-\ln x)^2 dx\).
02

Simplify the Integral

The integral simplifies to: \(V = \pi \int_1^{+\infty} (\ln x)^2 dx\).
03

Use Integration by Parts

Integration by parts is used here with \(u = \ln x\) and \(v = x\). The derivative of \(u\) (\(du\)) is \(1/x dx\) and the integral of \(v\) (\(dv\)) is \(dx\). The formula for integration by parts is \(\int udv = uv - \int vdu\). Using this, we compute the integral: \( \pi[(\ln x) x - \int x(1/x) dx]_1^{+\infty}\).
04

Calculate the Integral and Evaluate the Result

Evaluate the integral: \(= \pi[(\ln x) x - (x - 1)]_1^{+\infty}\). As \(x\) approaches infinity and \(x = 1\), calculate the result. Considering the property of the logarithm such as \(\ln 1 = 0\) and the fact that \(\ln x\) grows much slower than \(x\), we conclude that the first term in the bracket approaches 0 when \(x\) approaches infinity. The result is \(V = \pi \times 0 = 0\ cubic\ units.\)

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Most popular questions from this chapter

Given continuous functions \(f\) and \(g\) such that \(0 \leq f(x) \leq g(x)\) on the interval \([a, \infty),\) prove the following. (a) If \(\int_{a}^{\infty} g(x) d x\) converges, then \(\int_{a}^{\infty} f(x) d x\) converges. (b) If \(\int_{a}^{\infty} f(x) d x\) diverges, then \(\int_{a}^{\infty} g(x) d x\) diverges.

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