Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The solid is a rectangular prism with volume 18.

Step 1: Understand the Problem Statement

We are given a double integral over a rectangular region in the xy-plane, defined as \(R=\{(x, y)\) : \(0 \leq x \leq 2, 0 \leq y \leq 3\}\). The integral is \(\iint_{R} 3 \, dA\), which represents the volume of a solid.

Step 2: Interpret the Double Integral

A double integral \(\iint_{R} f(x, y) \, dA\) represents the volume under the surface \(z = f(x, y)\) over the region \(R\). Here, \(f(x, y) = 3\) indicates a constant function, meaning the surface is a horizontal plane at \(z = 3\).

Step 3: Identify the Base of the Solid

The base of the solid comes from region \(R\), which is a rectangle with vertices at \((0,0), (2,0), (2,3), (0,3)\). This rectangle covers the area from \(x = 0\) to \(x = 2\) and from \(y = 0\) to \(y = 3\).

Step 4: Determine the Shape of the Solid

Since \(f(x, y)=3\), the solid is a rectangular prism (a box) with height 3. The prism has its base on the xy-plane over the given rectangle \(R\), and extends uniformly up to \(z=3\).

Step 5: Sketch the Solid

Visualize the solid as a box with the rectangle \(R\) (2 units long and 3 units wide) as its base on the xy-plane and a constant height of 3 units. Thus, its other dimension, the height, simply extends to all points in \(R\) as \(z=3\).

Step 6: Compute the Volume

To find the volume of the solid, evaluate the double integral \(\iint_{R} 3 \, dA\). The area \(A\) of \(R\) is \(2\times3=6\). So, the integral evaluates to \(3 \times 6 = 18\).

Key Concepts

Volume Calculation

When we talk about volume calculation using double integrals, we refer to finding the volume of a solid that lies under a given surface over a specific region in the xy-plane. A double integral, like \( \iint_R f(x, y) \, dA \), helps compute this volume. Here, \( R \) denotes the region of interest, and \( f(x, y) \) is the function defining the surface height above each point on \( R \).
In the exercise, the function is \( f(x, y) = 3 \), indicating a constant function. This means the surface is a flat plane parallel to the xy-plane, precisely at a height of 3 units. Thus, the volume is calculated as the product of this constant height and the area of the base region \( R \). To simplify:

• Identify the area of \( R \).
• Multiply the area of \( R \) by the height, \( f(x, y) = 3 \).

This computation results in the total volume under the surface across the specified region.

Rectangular Region

A rectangular region in the context of a double integral is an area on the xy-plane where the function is evaluated. In our problem, the region \( R \) is defined by the rectangle with vertices at points \((0, 0), (2, 0), (2, 3), (0, 3)\).
This rectangle stretches from \( x = 0 \) to \( x = 2 \) horizontally and from \( y = 0 \) to \( y = 3 \) vertically. Such a clearly defined region helps in simplifying the computation of double integrals by providing neat limits for integration.
Think of a rectangular region as a flat surface bounded by straight lines parallel to the x and y-axes. This region sets the foundation upon which the entire problem of determining volume is built. Specifically, it provides the base on which the height of the solid rests.

Rectangular Prism

A rectangular prism is a three-dimensional shape with six faces where all angles are right angles, and opposite faces are equal. Imagine a box or a cuboid—you have a solid with rectangular faces all around.
In our exercise, the base of the rectangular prism is the established rectangular region \( R \) on the xy-plane. The function \( f(x, y) = 3 \) as a constant indicates that the prism’s height is 3 units.
This means that every point on the base stretches uniformly up to a height of 3, forming a rectangular prism. A rectangular prism's "box-like" shape simplifies volume computation, as it is merely the area of its base multiplied by its height. Knowing that helps us visualize the solid better, as it is essentially an extension of the region \( R \) into three dimensions.

Constant Function

A constant function, such as \( f(x, y) = 3 \) in this instance, is a function that gives the same output (3) for any input \( (x, y) \) in the domain. Such functions create surfaces that are flat planes parallel to one axis, in this case, the xy-plane.
The value of the constant function establishes the height at which this plane lies—here, \( z = 3 \). Consequently, this transforms the solid formed by the double integral into a box with consistent height across all points in the rectangular base \( R \).
Constant functions, while simple, provide significant simplification in volume calculations because they consolidate the integration process to just multiplying their value by the area of the region, making them a fundamental element in solving integrals over flat surfaces.