the region bounded by y = x^(-1/3), the x - axis, x = 2, and x = 4 is revolved about the x - axis. find the…

Explanation:

Step1: Recall volume - of - revolution 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 $a$ and $b$ are the limits of integration. Here, $f(x)=x^{-\frac{1}{3}}$, $a = 2$, and $b = 4$.

Step2: Set up the integral

Substitute the function and limits into the formula: $V=\pi\int_{2}^{4}(x^{-\frac{1}{3}})^{2}dx=\pi\int_{2}^{4}x^{-\frac{2}{3}}dx$.

Step3: Integrate the function

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int x^{-\frac{2}{3}}dx=\frac{x^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+C = 3x^{\frac{1}{3}}+C$.

Step4: Evaluate the definite integral

$V=\pi\left[3x^{\frac{1}{3}}\right]_{2}^{4}=\pi\left(3\times4^{\frac{1}{3}}-3\times2^{\frac{1}{3}}\right)=3\pi\left(2^{\frac{2}{3}}-2^{\frac{1}{3}}\right)$.

Answer:

$3\pi\left(2^{\frac{2}{3}}-2^{\frac{1}{3}}\right)$