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 solid is a 3D shape formed over a rectangle base in the XY-plane with heights from 1 to 3, increasing with \(x\).

Step 1: Understand the Region of Integration

The problem describes a rectangle defined by the bounds for \(x\) and \(y\). It specifically refers to the region \(R\) which is defined as \(0 \leq x \leq 2\) and \(0 \leq y \leq 3\). This gives a rectangular base in the XY-plane.

Step 2: Analyze the Given Function

The integrand in the double integral is \(x + 1\). This function represents the height of the solid above each point \((x, y)\) in the region \(R\). Since \(x\) ranges from 0 to 2, the height \(x+1\) ranges from 1 to 3 across this interval.

Step 3: Sketch the Base Region

Draw a rectangle in the XY-plane with vertices at \((0,0), (2,0), (2,3),\) and \((0,3)\). This rectangle represents the region \(R\) over which we will calculate the volume of the solid.

Step 4: Visualize the Solid

Thinking of \(x + 1\) as the height of the solid above each point in \(R\), start at \(x = 0\), where the height is 1, and increase to a height of 3 at \(x = 2\). This creates a slanted plane parallel to the Y-axis and linear from left to right, forming a 3D solid with a rectangular base.

Step 5: Sketch the Solid

The sketch of the solid should have a rectangular base located on the XY-plane with a varying height defined by \(z = x + 1\). Along the line from \(y = 0\) to \(y = 3\), the height stays constant for each fixed \(x\) from 0 to 2. The surface on top is a plane slanting upwards in the direction of increasing \(x\) values from 1 to 3.

Key Concepts

Volume of Solids

The volume of a solid is a measure of the space it occupies in three dimensions. When dealing with double integrals over a region, such as a rectangular area, we are often interested in finding the enclosed volume beneath a surface or above a given plane. In this exercise, the volume of the solid is captured by the double integral: \[ \iint_{R} (x + 1) \, dA \] Here, the function \( x + 1 \) represents the height of the solid at every point \((x, y)\) on the rectangular base. To imagine this, consider the base as a flat rectangle on the XY-plane and the height as extending vertically to form a slanted surface. The double integral accumulates an infinite number of infinitesimally small rectangular prisms over the area \( R \). The sum of these prisms gives you the volume of the solid.

Rectangular Region

A rectangular region is a simple, yet significant concept in double integrals. It refers to a subset of the plane bounded by two horizontal and two vertical sides. In this problem, the rectangular region \( R \) is defined by:

• \(0 \leq x \leq 2\)
• \(0 \leq y \leq 3\)

This creates a rectangle in the XY-plane with a width of 2 along the \(x\)-axis and a height of 3 along the \(y\)-axis. Picture this rectangle as a laying area on a flat surface where each point inside will be used to evaluate the height \(x + 1\). This foundation simplifies calculating the solid's overall volume using a double integral. Think of it as the canvas upon which the structure rises.

Integration Limits

Integration limits define the boundaries in which a function is evaluated. They are crucial for computational completeness in definite integrals. In double integration, these limits are set to evaluate along two dimensions.For our rectangular region \( R \):

• The limits for \(x\) are from 0 to 2.
• The limits for \(y\) are from 0 to 3.

These parameters guide the double integral: \[ \int_{0}^{2} \int_{0}^{3} (x+1) \, dy \, dx \]The integration proceeds first with respect to \(y\), holding \(x\) constant, and then with respect to \(x\). Each additional limit serves as a layer of specification, ensuring the function is systematically examined over the entire area of interest.

Surface Plotting

Surface plotting is a practical tool for visualizing the three-dimensional form of mathematical functions. By constructing a surface plot, you can see how a surface rises above a base region.In this exercise, plotting the surface \(z = x + 1\) over the rectangular region \( R \) allows you to observe the solid flesh out into space.

• Start by plotting the rectangle base on the XY-plane.
• Visualize how for each \(x\), the height \(z = x + 1\) is drawn parallel to the \(y\)-axis.
• See the slanting plane that extends from a height of 1 at \(x = 0\) to 3 at \(x = 2\).

This visual approach provides an immediate understanding of how the integration reflects the volume shaded beneath the surface, making abstract mathematical concepts more tangible.