Answer
\[V=2\pi \]
Work Step by Step
\[\begin{align}
& f\left( x \right)=-\ln x,\text{ }\left( 0,1 \right] \\
& \text{From the graph shown below, we can calculate the volume} \\
& \text{using the Disk Method about the }x\text{-Axis }V=\int_{a}^{b}{\pi f{{\left( x \right)}^{2}}dx,\text{ }} \\
& V=\int_{0}^{1}{\pi {{\left[ -\ln x \right]}^{2}}dx} \\
& V=\pi \int_{0}^{1}{{{\ln }^{2}}xdx} \\
& \text{The integrand is not defined for }x=0,\text{ then} \\
& V=\pi \underset{b\to {{0}^{+}}}{\mathop{\lim }}\,\int_{0}^{b}{{{\ln }^{2}}xdx} \\
& \text{Integrating }\int{{{\ln }^{2}}x}dx\text{ by parts} \\
& u={{\ln }^{2}}x\Rightarrow du=\frac{2\ln x}{x}dx \\
& dv=dx\Rightarrow v=x \\
& \int{{{\ln }^{2}}x}dx=x{{\ln }^{2}}x-\int{x\left( \frac{2\ln x}{x} \right)}dx \\
& \text{ }=x{{\ln }^{2}}x-2\int{\ln x}dx \\
& \text{Use a formula for }\int{\ln x}dx \\
& \text{ }=x{{\ln }^{2}}x-2\left( x\ln x-1 \right)+C \\
& \text{ }=x{{\ln }^{2}}x-2x\ln x+2x+C \\
& \text{Therefore,} \\
& V=\pi \underset{b\to {{0}^{+}}}{\mathop{\lim }}\,\left[ x{{\ln }^{2}}x-2x\ln x+2x \right]_{b}^{1} \\
& V=\pi \underset{b\to {{0}^{+}}}{\mathop{\lim }}\,\left[ 1{{\ln }^{2}}1-2\ln 1+2 \right]-\pi \underset{b\to {{0}^{+}}}{\mathop{\lim }}\,\left[ b{{\ln }^{2}}b-2b\ln b+2b \right] \\
& V=\pi \underset{b\to {{0}^{+}}}{\mathop{\lim }}\,\left[ 2 \right]-\pi \underset{b\to {{0}^{+}}}{\mathop{\lim }}\,\left[ b{{\ln }^{2}}b-2b\ln b+2b \right] \\
& \text{Evaluate the limit} \\
& V=\pi \left[ 2 \right]-\pi \left[ 0{{\ln }^{2}}0-2\left( 0 \right)\ln 0+2\left( 0 \right) \right] \\
& V=2\pi -\pi \left[ 0 \right] \\
& V=2\pi \\
\end{align}\]
