let r be the region bounded by y = ln x, the x - axis, the line x = e.\n(a) sketch the graphs and shade the…

Explanation:

Step1: Recall volume - disk method formula

The formula for the volume $V$ of a solid of revolution about the $x -$axis using the disk method is $V=\pi\int_{a}^{b}[f(x)]^{2}dx$, where $y = f(x)$ is the curve being revolved and $[a,b]$ is the interval of integration. Here, $y=\ln x$, $a = 1$ (since $\ln x$ intersects the $x -$axis at $x = 1$ where $\ln(1)=0$) and $b = e$.

Step2: Set up the integral

$V=\pi\int_{1}^{e}(\ln x)^{2}dx$.

Step3: Use integration by parts

The formula for integration by parts is $\int u;dv=uv-\int v;du$. Let $u = (\ln x)^{2}$ and $dv=dx$. Then $du=\frac{2\ln x}{x}dx$ and $v=x$. $V=\pi\left[x(\ln x)^{2}\big|{1}^{e}-\int{1}^{e}x\cdot\frac{2\ln x}{x}dx\right]=\pi\left[x(\ln x)^{2}\big|{1}^{e}- 2\int{1}^{e}\ln x;dx\right]$.

Step4: Use integration by parts again for $\int\ln x;dx$

Let $u=\ln x$ and $dv = dx$. Then $du=\frac{1}{x}dx$ and $v=x$. So $\int\ln x;dx=x\ln x-\int x\cdot\frac{1}{x}dx=x\ln x - x+C$. $V=\pi\left[x(\ln x)^{2}\big|{1}^{e}-2(x\ln x - x)\big|{1}^{e}\right]$.

Step5: Evaluate the definite - integrals

First, evaluate $x(\ln x)^{2}\big|{1}^{e}$: When $x = e$, $x(\ln x)^{2}=e(\ln e)^{2}=e\cdot1^{2}=e$. When $x = 1$, $x(\ln x)^{2}=1\cdot(\ln 1)^{2}=0$. Next, evaluate $2(x\ln x - x)\big|{1}^{e}$: When $x = e$, $2(x\ln x - x)=2(e\ln e - e)=2(e - e)=0$. When $x = 1$, $2(x\ln x - x)=2(1\cdot\ln 1-1)=2(0 - 1)=- 2$. $V=\pi\left[e-0-(0 + 2)\right]=\pi(e - 2)$.

Answer:

The volume of the solid of revolution is $\pi(e - 2)$