Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the solid whose volume is given by the following double integrals over the rectangle \(R=\\{(x, y)\) : \(0 \leq x \leq 2,0 \leq y \leq 3\\}\). \(\iint_{R}(x+1) d A\)

Step-by-Step Solution

Answer

The volume of the solid is 15 cubic units.

Step 1: Interpret the Double Integral

The given double integral \( \iint_{R} f(x, y) \ dA \) represents the volume under the surface \( z = f(x, y) \) over the region \( R \). Here, \( f(x, y) = x + 1 \). The double integral calculates the volume under \( z = x + 1 \) above the rectangle \( R = \{(x, y): 0 \leq x \leq 2, 0 \leq y \leq 3\} \).

Step 2: Visualize the Region R

The region \( R \) is a rectangle on the \( xy \)-plane, described by \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 3 \). This rectangle lies in the first quadrant starting from the origin, with its edges along the respective \( x \)-bound and \( y \)-bound.

Step 3: Identify the Surface Above the Region

The surface given by \( z = x + 1 \) is a plane that increases with \( x \). For each fixed \( y \), \( z \) varies from 1 to 3 as \( x \) varies from 0 to 2. This plane intersects the \( z \)-axis at \( z = 1 \) and is slanted upwards as \( x \) increases.

Step 4: Sketch the Solid

To sketch the solid, draw the rectangle \( R \) on the \( xy \)-plane. Then, draw the plane \( z = x + 1 \). At \( x = 0 \), \( z = 1 \), and at \( x = 2 \), \( z = 3 \). Draw lines vertically upwards from each point on \( R \) to this plane to form a trapezoidal prism-like shape. The solid is bounded by \( z = 1 \) when \( x = 0 \) to \( z = 3 \) when \( x = 2 \) across \( y \).

Step 5: Determine Bounds for the Integration

The integration bounds for the double integral are \( x \) from 0 to 2 and \( y \) from 0 to 3. The integral is set up as: \( \int_0^3 \int_0^2 (x + 1) \, dx \, dy \).

Step 6: Evaluate the Inner Integral

Evaluate the inner integral with respect to \( x \): \( \int_0^2 (x + 1) \, dx \). This yields: \( \left[ \frac{x^2}{2} + x \right]_0^2 = \left( \frac{2^2}{2} + 2 \right) - \left( 0 + 0 \right) = 3 + 2 = 5 \).

Step 7: Evaluate the Outer Integral

With the result from Step 6, evaluate the outer integral with respect to \( y \): \( \int_0^3 5 \, dy \). This yields: \( \left[ 5y \right]_0^3 = 5(3) - 5(0) = 15 \).

Step 8: Final Result

The volume of the solid is the value of the double integral, which is 15 cubic units.

Key Concepts

Volume Calculation

In the context of double integrals, volume calculation involves determining the amount of space beneath a surface that lies above a certain region in the plane. To achieve this, we perform a double integral over a specific area under the function, which denotes the height of our solid above each point in the region.
Double integrals essentially "accumulate" small volumes across our specified area, where each volume segment is the function value (or height) multiplied by a tiny area segment.
In our original exercise, the double integral \( \iint_{R}(x+1) \, dA \) determines the volume under the surface described by \( z = x + 1 \) above the rectangle \( R \).
With our setup, we integrate the function \( x + 1 \) over the rectangular region \( R \) to compute this total volume.

Surface Planes

The term "surface plane" in this context refers to the mathematical function that represents a continuous, flat surface.
In our exercise, the surface plane is described by the equation \( z = x + 1 \). This plane ascends linearly as \( x \) increases, representing a diagonal slope on the z-axis.
For each constant \( y \), the surface height increases linearly from \( z = 1 \) when \( x = 0 \) to \( z = 3 \) when \( x = 2 \). Given the function \( z = x + 1 \), this surface plane is one continuous, tilted sheet originating from the xy-plane.
Its inclination is uniform because it is a linear function—essentially, it has a slope of 1 along the x-direction. Understanding this planar surface gives us insight into why the volume calculation results in a three-dimensional trapezoidal-like shape.

Integration Bounds

Integration bounds are critical as they define the limits of our region of interest for the double integral calculation.
For the given problem, the rectangular region \( R = \{(x, y): 0 \leq x \leq 2, 0 \leq y \leq 3\} \) forms our area of calculation on the xy-plane.
When setting up the double integral, the integration bounds are determined as follows:

• The inner integral, with respect to \( x \), spans from 0 to 2.
• The outer integral, with respect to \( y \), spans from 0 to 3.

These bounds ensure that we only consider the volume under the surface \( z = x + 1 \) and directly above the rectangle \( R \). By integrating sequentially over these bounds, we systematically cover the entire specified region, capturing all the infinitesimal sections that comprise the solid's total volume.

Rectangular Regions

Rectangular regions provide a simplified means of defining integration limits in double integrals. The rectangular region for this problem is denoted by \( R = \{(x, y): 0 \leq x \leq 2, 0 \leq y \leq 3\} \).
This rectangle is composed of all points \((x,y)\) where \(x\) ranges between 0 and 2, and \(y\) ranges between 0 and 3.
On a standard xy-plane, this forms a compact area with horizontal and vertical boundaries. Such a region means our surface calculations are confined to a simple, concrete shape.

• This alignment aids in establishing a stable groundwork for integration as the calculations can navigate straightforward upper and lower limits.
• Due to its geometric simplicity, it is easier to visualize and analyze the volume located directly above it.

When dealing with rectangular regions, integration over it becomes systematic—first along one axis and then the other—leading to accurate and reliable calculations of areas and volumes.