let r be the region bounded by the following curves. find the volume of the solid generated when r is…

Explanation:

Step1: Recall the disk - method formula

The formula for the volume $V$ of the 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 function and $[a,b]$ is the interval of integration. Here, $f(x)=\frac{4}{\sqrt{x}}$, $a = 2$, and $b = 4$.

Step2: Square the function

[ \begin{align*} [f(x)]^{2}&=\left(\frac{4}{\sqrt{x}}\right)^{2}\ &=\frac{16}{x} \end{align*} ]

Step3: Set up the integral

The volume $V=\pi\int_{2}^{4}\frac{16}{x}dx$.

Step4: Integrate

We know that $\int\frac{1}{x}dx=\ln|x|+C$. So, $\int_{2}^{4}\frac{16}{x}dx=16\int_{2}^{4}\frac{1}{x}dx$. [ \begin{align*} 16\int_{2}^{4}\frac{1}{x}dx&=16[\ln x]_{2}^{4}\ &=16(\ln4-\ln2) \end{align*} ] Using the property of logarithms $\ln a-\ln b=\ln\frac{a}{b}$, we have $16(\ln4 - \ln2)=16\ln\frac{4}{2}=16\ln2$.

Step5: Find the volume

Since $V=\pi\int_{2}^{4}\frac{16}{x}dx$, then $V = 16\pi\ln2$.

Answer:

$16\pi\ln2$