The volume of the solid is found by integrating the area of the cross-sections along the $x$-axis.

Step 1: Understanding the Cross Sections

Each cross-section perpendicular to the $x$-axis is a square with side length equal to the height of the region $R$. The height of the region is given by:

$\text{Height} = f(x) - (-1) = f(x) + 1$

Since the given function is $f(x) = \sqrt{x} + 2$, we have:

$\text{Height} = (\sqrt{x} + 2) + 1 = \sqrt{x} + 3$

The area of a square cross-section is:

$A(x) = (\text{side length})^2 = (\sqrt{x} + 3)^2$

Step 2: Setting Up the Integral

The volume of the solid is obtained by integrating the cross-sectional area from $x = 0$ to $x = 6$:

$V = \int_0^6 (\sqrt{x} + 3)^2 \, dx$

Expanding the square:

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

Thus, the integral simplifies to:

$V = \int_0^6 (x + 6\sqrt{x} + 9) \, dx$

Step 3: Evaluating the Integral

We evaluate each term separately:

1. Integral of $x$: $\int x \, dx = \frac{x^2}{2}$ Evaluated from 0 to 6: $\frac{6^2}{2} - \frac{0^2}{2} = \frac{36}{2} = 18$

2. Integral of $6\sqrt{x}$: $\int 6x^{1/2} \, dx = 6 \cdot \frac{2}{3} x^{3/2} = 4x^{3/2}$ Evaluated from 0 to 6: $4(6^{3/2}) - 4(0^{3/2})$ Since $6^{3/2} = 6 \times \sqrt{6} \approx 6 \times 2.449 = 14.694$:

   $4(14.694) = 58.776$

3. Integral of $9$: $\int 9 \, dx = 9x$ Evaluated from 0 to 6: $9(6) - 9(0) = 54$

Step 4: Summing the Results

$V = 18 + 58.776 + 54 = 130.776$

Final Answer

$V \approx 130.776 \quad \text{(rounded to the nearest thousandth)}$

Would you like a breakdown of any specific step?

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Related Problems:

1. What would the volume be if the cross-sections were semicircles instead of squares?
2. How does the volume change if the upper function were $f(x) = x + 3$ instead?
3. What happens to the volume if the limits of integration change to $x = 2$ to $x = 6$?
4. What if the cross-sections were equilateral triangles instead of squares?
5. How do you compute the volume if the base region were rotated around the $x$-axis instead?

Tip: When dealing with cross-sectional volume problems, always determine the side length of the cross-section first and square it before integrating!