let the region r be the area enclosed by the function f(x)=ln(x) and g(x)=3/2x - 2. find the volume of the…

Explanation:

Step1: Find intersection points

Set $\ln(x)=\frac{3}{2}x - 2$. Using a calculator or software, the intersection points are $x = 1$ and $x\approx1.513$.

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, $R(x)=\frac{3}{2}x - 2+3=\frac{3}{2}x + 1$ and $r(x)=\ln(x)+3$, and $a = 1$, $b\approx1.513$. So $V=\pi\int_{1}^{1.513}[(\frac{3}{2}x + 1)^{2}-(\ln(x)+3)^{2}]dx$.

Step3: Expand the integrand

Expand $(\frac{3}{2}x + 1)^{2}=\frac{9}{4}x^{2}+3x + 1$ and $(\ln(x)+3)^{2}=\ln^{2}(x)+6\ln(x)+9$. The integrand becomes $\frac{9}{4}x^{2}+3x + 1-(\ln^{2}(x)+6\ln(x)+9)=\frac{9}{4}x^{2}+3x-\ln^{2}(x)-6\ln(x)-8$.

Step4: Integrate term - by - term

$\int\frac{9}{4}x^{2}dx=\frac{9}{4}\times\frac{1}{3}x^{3}=\frac{3}{4}x^{3}$, $\int3xdx=\frac{3}{2}x^{2}$, $\int\ln^{2}(x)dx=x\ln^{2}(x)-2x\ln(x)+2x$ (using integration by parts twice), $\int\ln(x)dx=x\ln(x)-x$, $\int - 8dx=-8x$.

Step5: Evaluate the definite integral

$V=\pi\left[\frac{3}{4}x^{3}+\frac{3}{2}x^{2}-(x\ln^{2}(x)-2x\ln(x)+2x)-6(x\ln(x)-x)-8x\right]_{1}^{1.513}$. Using a calculator to evaluate the above expression: $V\approx1.077$.

Answer:

$1.077$