Question

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

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

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: \(V = \pi \int_1^{+\infty} (-\ln x)^2 dx\).

Simplify the Integral

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

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: \( \pi[(\ln x) x - \int x(1/x) dx]_1^{+\infty}\).

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 \(V = \pi \times 0 = 0\ cubic\ units.\)