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

Explanation:

Step1: Determine the outer - radius and inner - radius functions

The outer - radius $R(x)$ is the distance from the line $y = 6$ to the lower - bound $y=-1$, so $R(x)=6 - (-1)=7$. The inner - radius $r(x)$ is the distance from the line $y = 6$ to the function $y=\sqrt{x}+2$, so $r(x)=6-(\sqrt{x}+2)=4 - \sqrt{x}$.

Step2: Use the washer method formula

The volume $V$ of the solid of revolution using the washer method is given by $V=\pi\int_{a}^{b}(R^{2}(x)-r^{2}(x))dx$, where $a = 0$, $b = 6$. So $V=\pi\int_{0}^{6}(7^{2}-(4 - \sqrt{x})^{2})dx$.

Step3: Expand the integrand

Expand $(4 - \sqrt{x})^{2}=16-8\sqrt{x}+x$. Then $7^{2}-(4 - \sqrt{x})^{2}=49-(16 - 8\sqrt{x}+x)=33 + 8\sqrt{x}-x$.

Step4: Integrate term - by - term

$\int(33 + 8\sqrt{x}-x)dx=33x+8\times\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{2}x^{2}+C=33x+\frac{16}{3}x^{\frac{3}{2}}-\frac{1}{2}x^{2}+C$.

Step5: Evaluate the definite integral

$V=\pi\left[33x+\frac{16}{3}x^{\frac{3}{2}}-\frac{1}{2}x^{2}\right]_{0}^{6}=\pi\left(33\times6+\frac{16}{3}\times6^{\frac{3}{2}}-\frac{1}{2}\times6^{2}\right)$. $33\times6 = 198$, $\frac{16}{3}\times6^{\frac{3}{2}}=\frac{16}{3}\times6\sqrt{6}=32\sqrt{6}$, $\frac{1}{2}\times6^{2}=18$. $V=\pi(198 + 32\sqrt{6}-18)=\pi(180 + 32\sqrt{6})\approx\pi(180+32\times2.44949)\approx\pi(180 + 78.3837)\approx\pi\times258.3837\approx811.797$.

Answer:

$811.797$