the region bounded by $y = x^{-\frac{1}{4}}$, the $x$-axis, $x = 1$, and $x = 4$ is revolved about the…

Explanation:

Step1: Apply the shell - method formula

The shell - method formula for the volume $V$ of a solid of revolution about the $y$-axis is $V = 2\pi\int_{a}^{b}x\cdot h(x)dx$, where $a$ and $b$ are the limits of integration, $x$ is the radius of the shell, and $h(x)$ is the height of the shell. Here, $a = 1$, $b = 4$, and $h(x)=x^{-\frac{1}{4}}$. So, $V = 2\pi\int_{1}^{4}x\cdot x^{-\frac{1}{4}}dx$.

Step2: Simplify the integrand

Using the rule of exponents $x^m\cdot x^n=x^{m + n}$, we have $x\cdot x^{-\frac{1}{4}}=x^{1-\frac{1}{4}}=x^{\frac{3}{4}}$. So the integral becomes $V = 2\pi\int_{1}^{4}x^{\frac{3}{4}}dx$.

Step3: Integrate the function

The antiderivative of $x^{\frac{3}{4}}$ is $\frac{4}{7}x^{\frac{7}{4}}$ (using the power - rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$, where $n=\frac{3}{4}$). Then $V = 2\pi\left[\frac{4}{7}x^{\frac{7}{4}}\right]_{1}^{4}$.

Step4: Evaluate the definite integral

First, substitute $x = 4$ and $x = 1$ into $\frac{4}{7}x^{\frac{7}{4}}$. When $x = 4$, $\frac{4}{7}x^{\frac{7}{4}}=\frac{4}{7}(4)^{\frac{7}{4}}=\frac{4}{7}(2^2)^{\frac{7}{4}}=\frac{4}{7}\times2^{\frac{7}{2}}=\frac{4}{7}\times2^3\times2^{\frac{1}{2}}=\frac{32}{7}\sqrt{2}$. When $x = 1$, $\frac{4}{7}x^{\frac{7}{4}}=\frac{4}{7}(1)^{\frac{7}{4}}=\frac{4}{7}$. Then $V = 2\pi\left(\frac{32}{7}\sqrt{2}-\frac{4}{7}\right)=\frac{8\pi}{7}(8\sqrt{2}-1)$.

Answer:

$\frac{8\pi}{7}(8\sqrt{2}-1)$