let the region r be the area enclosed by the function f(x) = √x - 2, the horizontal line y = -1, and the y…

Explanation:

Step1: Find the intersection point

Set $\sqrt{x}-2=-1$. Solving for $x$ gives $\sqrt{x}=1$, so $x = 1$.

Step2: Use the disk - washer method formula

The formula for the volume $V$ of a solid of revolution about the line $y = k$ using the disk - washer method is $V=\pi\int_{a}^{b}[(R(x))^{2}]dx$, where $R(x)$ is the radius of the cross - sectional disk. Here, $a = 0$, $b = 1$, and $R(x)=(\sqrt{x}-2 + 1)=(\sqrt{x}-1)$. So $V=\pi\int_{0}^{1}(\sqrt{x}-1)^{2}dx$.

Step3: Expand the integrand

Expand $(\sqrt{x}-1)^{2}=x - 2\sqrt{x}+1$. Then $V=\pi\int_{0}^{1}(x - 2x^{\frac{1}{2}}+1)dx$.

Step4: Integrate term - by - term

$\int(x - 2x^{\frac{1}{2}}+1)dx=\frac{1}{2}x^{2}-2\times\frac{2}{3}x^{\frac{3}{2}}+x+C=\frac{1}{2}x^{2}-\frac{4}{3}x^{\frac{3}{2}}+x+C$.

Step5: Evaluate the definite integral

$V=\pi\left[\frac{1}{2}x^{2}-\frac{4}{3}x^{\frac{3}{2}}+x\right]_{0}^{1}=\pi\left(\frac{1}{2}-\frac{4}{3}+1\right)$.

Step6: Simplify the result

$V=\pi\left(\frac{3 - 8 + 6}{6}\right)=\frac{\pi}{6}\approx0.524$.

Answer:

$0.524$