question let the region r be the area enclosed by the function f(x)=ln(x)+1 and g(x)=x - 1. find the volume…

Explanation:

Step1: Find intersection points

Set $\ln(x)+1=x - 1$. Using a calculator or numerical methods, the intersection points are $x_1\approx0.159$ and $x_2\approx3.146$.

Step2: Use the washer - method formula

The formula for the volume $V$ of the solid of revolution about the line $y = k$ using the washer - method is $V=\pi\int_{a}^{b}([R(x)]^{2}-[r(x)]^{2})dx$, where $R(x)$ is the outer radius and $r(x)$ is the inner radius. Here, $R(x)=(x - 1)-(-2)=x + 1$ and $r(x)=(\ln(x)+1)-(-2)=\ln(x)+3$, and $a\approx0.159$, $b\approx3.146$. So $V=\pi\int_{0.159}^{3.146}((x + 1)^{2}-(\ln(x)+3)^{2})dx$.

Step3: Expand the integrand

Expand $(x + 1)^{2}-(\ln(x)+3)^{2}=x^{2}+2x + 1-(\ln^{2}(x)+6\ln(x)+9)=x^{2}+2x-\ln^{2}(x)-6\ln(x)-8$.

Step4: Integrate term - by - term

$\int x^{2}dx=\frac{1}{3}x^{3}$, $\int 2xdx=x^{2}$, $\int\ln^{2}(x)dx=x\ln^{2}(x)-2x\ln(x)+2x$, $\int\ln(x)dx=x\ln(x)-x$, $\int 8dx = 8x$. Then $\int(x^{2}+2x-\ln^{2}(x)-6\ln(x)-8)dx=\frac{1}{3}x^{3}+x^{2}-x\ln^{2}(x)+2x\ln(x)-2x-6(x\ln(x)-x)-8x=\frac{1}{3}x^{3}+x^{2}-x\ln^{2}(x)-4x\ln(x)-4x$.

Step5: Evaluate the definite integral

$V=\pi\left[\frac{1}{3}x^{3}+x^{2}-x\ln^{2}(x)-4x\ln(x)-4x\right]_{0.159}^{3.146}$. Using a calculator to evaluate the definite - integral and multiply by $\pi$, we get $V\approx36.968$.

Answer:

$36.968$