91Ó°ÊÓ

The double integral in this problem provides a way to find the volume of a solid. This solid is situated under the surface defined by the equation \( z = 2x + 1 \) over a specified region in the \( xy \)-plane. To visualize this, think of the region \( R \) as the base lying flat on a surface. Above this base, the surface \( z = 2x + 1 \) rises, creating a three-dimensional shape.

Calculating the double integral allows us to determine how much space this shape occupies, which is essentially the volume. When you set up a double integral, you're finding the accumulated 'height' (given by \( 2x + 1 \)) over every infinitesimal part of the region \( R \).

When done over all of \( R \), we get the total volume, which in this case is calculated to be 24 through the double integration process.

The region of integration, denoted by \( R \), is crucial for setting up a double integral. It defines the limits over which you'll be integrating. In this problem, \( R \) is the rectangle in the \( xy \)-plane with boundaries \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 4 \).

Imagine drawing the rectangle on a graph—it stretches horizontally from \( x = 0 \) to \( x = 2 \) and vertically from \( y = 0 \) to \( y = 4 \).

Identifying this region correctly is essential because it ensures that every part of the volume under the surface \( z = 2x + 1 \) within \( R \) is accounted for in your double integral.

The first step in evaluating this double integral is to integrate with respect to \( x \). In essence, you are calculating the contribution to the volume from each vertical slice parallel to the \( y \)-axis.

The function \( 2x + 1 \) is integrated over the limits from \( x = 0 \) to \( x = 2 \). This involves finding the antiderivative of \( 2x + 1 \), which is \( x^2 + x \). Apply these antiderivatives to the limits:\[\int_{0}^{2} (2x + 1) \, dx = \left[ x^2 + x \right]_{0}^{2} = 4 + 2 = 6\]

The result, 6, from this step acts as a simplified version of the problem, making it easier to perform the second integration step.

After integrating with respect to \( x \), the process continues with integrating the resulting expression, which is a constant 6, with respect to \( y \). This step measures how the summed contributions from the \( x \)-slices extend and accumulate along the \( y \)-dimension within the region.

The limits here are \( y = 0 \) to \( y = 4 \), and the calculation is straightforward: \[\int_{0}^{4} 6 \, dy = 6y \big|_{0}^{4} = 6 \times 4 - 6 \times 0 = 24\]
This result, 24, represents the total volume we were seeking to calculate. It confirms how integration first in \( x \) then in \( y \) accumulates to give the exact volume under the specified surface over the given region.