To set up the integral for the solid whose cross-section is perpendicular to the x-axis and whose side lies on the base, let’s break it down step by step.

Step 1: Understand the problem

The region $R$ is bounded by:

• $x = 4$
• $y = -2$
• $y = \ln(x)$

You are interested in finding the volume of the solid whose cross-sections perpendicular to the x-axis have sides lying along the base of this region.

Step 2: Find the limits of integration

The region $R$ is defined between $x = 1$ and $x = 4$ because $y = \ln(x)$ is only valid for $x > 0$ and we know it intersects $y = -2$ at $x = 1$ since $\ln(1) = 0$. Hence, the limits for $x$ are from 1 to 4.

Step 3: Describe the cross-sectional area

For each $x$, the length of the side of the cross-section is the vertical distance between the curves $y = -2$ and $y = \ln(x)$. The length of this side is:

$\text{Side length} = \ln(x) - (-2) = \ln(x) + 2$

Since the cross-sections are perpendicular to the x-axis, the area of each cross-section is a square with side length $\ln(x) + 2$.

Thus, the area of the cross-section at any point $x$ is:

$A(x) = \left( \ln(x) + 2 \right)^2$

Step 4: Set up the integral

To find the volume of the solid, integrate the area of the cross-sections from $x = 1$ to $x = 4$:

$V = \int_1^4 \left( \ln(x) + 2 \right)^2 \, dx$

Step 5: Simplify the integral expression

Expand the square to make the integration process easier:

$\left( \ln(x) + 2 \right)^2 = \ln^2(x) + 4\ln(x) + 4$

Thus, the volume integral becomes:

$V = \int_1^4 \left( \ln^2(x) + 4\ln(x) + 4 \right) \, dx$

Now, this is ready for evaluation!

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Would you like me to evaluate this integral for you, or do you have any specific questions about the process?

Here are some related questions to explore further:

1. How do you evaluate integrals involving logarithmic functions?
2. What is the physical interpretation of cross-sectional areas in this context?
3. How does changing the function for the base affect the volume?
4. How do you handle integrals of higher powers of logarithms?
5. What is the relationship between the geometry of the solid and the shape of the cross-sections?

Tip: When dealing with integrals involving logarithms, remember that the derivative of $\ln(x)$ is $1/x$, which can be useful for simplifying integrals.