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

Explanation:

Step1: Find the side - length of the square cross - section

The upper function is $y_1=\sqrt{x}+2$ and the lower function is $y_2 = - 1$. The side - length $s$ of the square cross - section perpendicular to the $x$ - axis is $s=(\sqrt{x}+2)-(-1)=\sqrt{x}+3$.

Step2: Set up the volume formula

The volume $V$ of the solid with square cross - sections perpendicular to the $x$ - axis over the interval $[a,b]$ is given by $V=\int_{a}^{b}s^{2}dx$. Here, $a = 0$, $b = 6$, and $s=\sqrt{x}+3$, so $V=\int_{0}^{6}(\sqrt{x}+3)^{2}dx$.

Step3: Expand the integrand

Expand $(\sqrt{x}+3)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$. We have $(\sqrt{x}+3)^{2}=x + 6\sqrt{x}+9$.

Step4: Integrate term - by - term

$\int_{0}^{6}(x + 6\sqrt{x}+9)dx=\int_{0}^{6}x dx+6\int_{0}^{6}x^{\frac{1}{2}}dx+\int_{0}^{6}9dx$. Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we get: $\int_{0}^{6}x dx=\left[\frac{x^{2}}{2}\right]{0}^{6}=\frac{6^{2}}{2}-\frac{0^{2}}{2}=18$; $6\int{0}^{6}x^{\frac{1}{2}}dx=6\left[\frac{2}{3}x^{\frac{3}{2}}\right]{0}^{6}=4x^{\frac{3}{2}}\big|{0}^{6}=4\times6^{\frac{3}{2}}=4\times6\sqrt{6}=24\sqrt{6}$; $\int_{0}^{6}9dx=9x\big|_{0}^{6}=9\times6 - 9\times0 = 54$.

Step5: Calculate the volume

$V=18 + 24\sqrt{6}+54=72+24\sqrt{6}\approx72+24\times2.44949\approx72 + 58.788=130.788$.

Answer:

$130.788$