region $r$ is enclosed by the lines $y = 1,x = 4$, and the curve $y=sqrt{x}$. what is the volume of the…

Explanation:

Step1: Recall the washer - method formula

The formula for the volume $V$ of the solid of revolution about the $x$-axis 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. For the region bounded by $y = \sqrt{x}$, $y = 1$, and $x = 4$, when rotating about the $x$-axis, the outer - radius $R(x)=\sqrt{x}$ and the inner - radius $r(x)=1$ for $x$ ranging from $1$ to $4$.

Step2: Set up the integral

We have $V=\pi\int_{1}^{4}((\sqrt{x})^{2}-1^{2})dx=\pi\int_{1}^{4}(x - 1)dx$.

Step3: Integrate term - by - term

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

Step4: Evaluate the definite integral

$V=\pi\left[\frac{x^{2}}{2}-x\right]_{1}^{4}=\pi\left(\left(\frac{4^{2}}{2}-4\right)-\left(\frac{1^{2}}{2}-1\right)\right)$. First, calculate $\frac{4^{2}}{2}-4=\frac{16}{2}-4=8 - 4 = 4$. Second, calculate $\frac{1^{2}}{2}-1=\frac{1}{2}-1=-\frac{1}{2}$. Then $V=\pi\left(4-\left(-\frac{1}{2}\right)\right)=\pi\left(4+\frac{1}{2}\right)=\frac{9\pi}{2}$.

Answer:

$\frac{9\pi}{2}$