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 $\sqrt{x}=3$, so $x = 9$.

Step2: Use the disk - washer method

The radius of the cross - section is $r(x)=1-(\sqrt{x}-2)=3 - \sqrt{x}$. The volume formula for rotating about a horizontal line $y = k$ using the disk method is $V=\pi\int_{a}^{b}[r(x)]^{2}dx$. Here, $a = 0$, $b = 9$, and $r(x)=3-\sqrt{x}$. So $V=\pi\int_{0}^{9}(3 - \sqrt{x})^{2}dx$.

Step3: Expand the integrand

Expand $(3 - \sqrt{x})^{2}=9-6\sqrt{x}+x$.

Step4: Integrate term - by - term

$\int_{0}^{9}(9-6\sqrt{x}+x)dx=\int_{0}^{9}9dx-6\int_{0}^{9}x^{\frac{1}{2}}dx+\int_{0}^{9}xdx$. $\int_{0}^{9}9dx=9x\big|{0}^{9}=81$. $6\int{0}^{9}x^{\frac{1}{2}}dx=6\times\frac{2}{3}x^{\frac{3}{2}}\big|{0}^{9}=4\times9^{\frac{3}{2}}=4\times27 = 108$. $\int{0}^{9}xdx=\frac{1}{2}x^{2}\big|{0}^{9}=\frac{81}{2}$. Then $\int{0}^{9}(9 - 6\sqrt{x}+x)dx=81-108+\frac{81}{2}=\frac{162 - 216+81}{2}=\frac{243 - 216}{2}=\frac{27}{2}$.

Step5: Calculate the volume

$V=\pi\times\frac{27}{2}\approx42.412$.

Answer:

$42.412$