let the region r be the area enclosed by the function f(x) = √x - 2, the x - axis, and the y - axis. find…

Explanation:

Step1: Recall the disk - method formula

The volume $V$ of the solid of revolution about the $x$-axis using the disk - method is $V=\pi\int_{a}^{b}[f(x)]^{2}dx$. First, we need to find the $x$-intercept of $y = \sqrt{x}-2$. Set $y = 0$, then $\sqrt{x}-2=0$, so $\sqrt{x}=2$ and $x = 4$. The region is bounded by $x = 0$ and $x = 4$, and $f(x)=\sqrt{x}-2$.

Step2: Expand the integrand

[ \begin{align*} [f(x)]^{2}&=(\sqrt{x}-2)^{2}\ &=x - 4\sqrt{x}+4 \end{align*} ]

Step3: Set up the integral

The volume $V=\pi\int_{0}^{4}(x - 4\sqrt{x}+4)dx$.

Step4: Integrate term - by - term

We know that $\int xdx=\frac{1}{2}x^{2}+C$, $\int\sqrt{x}dx=\int x^{\frac{1}{2}}dx=\frac{2}{3}x^{\frac{3}{2}}+C$, and $\int 4dx = 4x+C$. [ \begin{align*} V&=\pi\left[\frac{1}{2}x^{2}-4\times\frac{2}{3}x^{\frac{3}{2}} + 4x\right]_{0}^{4}\ &=\pi\left(\frac{1}{2}(4)^{2}-\frac{8}{3}(4)^{\frac{3}{2}}+4\times4\right)\ &=\pi\left(8-\frac{8}{3}\times8 + 16\right)\ &=\pi\left(8-\frac{64}{3}+16\right)\ &=\pi\left(\frac{24 - 64+48}{3}\right)\ &=\frac{8\pi}{3}\approx8.378 \end{align*} ]

Answer:

$8.378$