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Explanation:

Step1: Recall disk - method for volume about x - axis

The formula for the volume $V$ of the solid obtained by revolving the curve $y = f(x)$ from $x=a$ to $x = b$ about the $x$-axis is $V=\pi\int_{a}^{b}[f(x)]^{2}dx$. Here, $f(x)=\ln x$, $a = 1$, and $b = 8$. So $V=\pi\int_{1}^{8}(\ln x)^{2}dx$. We use integration by parts. Let $u = (\ln x)^{2}$ and $dv=dx$. Then $du=\frac{2\ln x}{x}dx$ and $v=x$. By the integration - by - parts formula $\int u;dv=uv-\int v;du$, we have $\int(\ln x)^{2}dx=x(\ln x)^{2}-2\int\ln xdx$. For $\int\ln xdx$, use integration by parts again. Let $u=\ln x$ and $dv = dx$, then $du=\frac{1}{x}dx$ and $v=x$, so $\int\ln xdx=x\ln x - x+C$. So $\int(\ln x)^{2}dx=x(\ln x)^{2}-2(x\ln x - x)+C$. Evaluating $\pi\int_{1}^{8}(\ln x)^{2}dx=\pi\left[x(\ln x)^{2}-2x\ln x + 2x\right]_{1}^{8}=\pi\left(8(\ln 8)^{2}-16\ln 8 + 16-(0 - 0+2)\right)=\pi\left(8(\ln 8)^{2}-16\ln 8 + 14\right)$.

Step2: Recall shell - method for volume about y - axis

The formula for the volume $V$ of the solid obtained by revolving the curve $y = f(x)$ from $x=a$ to $x = b$ about the $y$-axis using the shell method is $V = 2\pi\int_{a}^{b}x\cdot f(x)dx$. Here, $f(x)=\ln x$, $a = 1$, and $b = 8$. So $V=2\pi\int_{1}^{8}x\ln xdx$. Use integration by parts. Let $u=\ln x$ and $dv=xdx$. Then $du=\frac{1}{x}dx$ and $v=\frac{1}{2}x^{2}$. By the integration - by - parts formula $\int u;dv=uv-\int v;du$, we have $\int x\ln xdx=\frac{1}{2}x^{2}\ln x-\frac{1}{2}\int xdx=\frac{1}{2}x^{2}\ln x-\frac{1}{4}x^{2}+C$. Evaluating $2\pi\int_{1}^{8}x\ln xdx=2\pi\left[\frac{1}{2}x^{2}\ln x-\frac{1}{4}x^{2}\right]_{1}^{8}=2\pi\left(\frac{1}{2}\times64\ln 8-\frac{1}{4}\times64-\left(0-\frac{1}{4}\right)\right)=2\pi\left(32\ln 8 - 16+\frac{1}{4}\right)=2\pi\left(32\ln 8-\frac{63}{4}\right)$.

Answer:

(a) $\pi\left(8(\ln 8)^{2}-16\ln 8 + 14\right)$ (b) $2\pi\left(32\ln 8-\frac{63}{4}\right)$