Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the comparison property of double integrals to show that if \(f(x, y) \geq 0\) on \(R\) then \(\iint_{R} f(x, y) d A \geq 0\).

Step-by-Step Solution

Answer

Since \( f(x, y) \geq 0 \) on \( R \), \( \iint_{R} f(x, y) \, dA \geq 0 \) by the comparison property of integrals.

Step 1: Understand the Domain and Function

We are given a function \( f(x, y) \) that is non-negative over a region \( R \). This means for every point \( (x, y) \) in \( R \), \( f(x, y) \geq 0 \).

Step 2: Recall the Comparison Property of Double Integrals

The comparison property of double integrals states that if \( f(x, y) \geq g(x, y) \) on a region \( R \), then \( \iint_{R} f(x, y) \, dA \geq \iint_{R} g(x, y) \, dA \). In this case, we compare \( f(x, y) \) to the function \( 0 \), since \( f(x, y) \geq 0 \).

Step 3: Apply the Comparison Property

By the comparison property, since \( f(x, y) \geq 0 \) on \( R \), we have \( \iint_{R} f(x, y) \, dA \geq \iint_{R} 0 \, dA \).

Step 4: Evaluate the Integral of 0 over R

The integral of 0 over any region \( R \) is 0: \( \iint_{R} 0 \, dA = 0 \).

Step 5: Conclude the Result

Putting it all together, since \( \iint_{R} f(x, y) \, dA \geq \iint_{R} 0 \, dA = 0 \), it follows that \( \iint_{R} f(x, y) \, dA \geq 0 \).

Key Concepts

Comparison Property

The comparison property of double integrals is a useful tool when analyzing functions over a given region. It helps to compare the integrals of two functions based on their pointwise relation within the defined region.
This property states that if you have two functions, say \( f(x, y) \) and \( g(x, y) \), and if \( f(x, y) \) is greater than or equal to \( g(x, y) \) for every point in a region \( R \), then their integrals will also maintain this inequality:

• \( \iint_{R} f(x, y) \, dA \geq \iint_{R} g(x, y) \, dA \)

Applying this property helps in establishing bounds and verifying certain inequalities for double integrals. It is particularly useful when one of the functions, such as in our exercise, is zero, as it simplifies the comparison.

Non-Negative Function

A non-negative function is one that does not take any negative values within its domain. In the context of our exercise, the function \( f(x, y) \) is said to be non-negative over the region \( R \), which mathematically can be expressed as \( f(x, y) \geq 0 \) for all points \( (x, y) \) in \( R \).
This property ensures that the function only contributes positive values (or zero) to the integral, implying a positive (or zero) value for the entire integral over this region. Here are some key points on non-negative functions:

• They are bound to result in a non-negative area under the curve when integrated over a region.
• They simplify many mathematical analyses involving inequalities and integral properties.
• Non-negative is a necessary condition in many physical applications such as probability density functions where negative values would not make sense.

Ensuring a function is non-negative can be a decisive aspect before applying certain integral properties like the comparison property.

Region of Integration

The region of integration \( R \) is the specific domain over which a double integral is evaluated. In practical terms, it can be visualized as the 'area' on the \( xy \)-plane that is bounded by certain constraints or functions. Understanding the region of integration is crucial because it determines the limits of integration and ultimately, the outcome of the integral itself. Here are some insights:

• The boundaries of \( R \) define where the function is evaluated.
• Choosing \( R \) can influence the ease or difficulty of computing the integral, often one looks for simple shapes like rectangles or circles that make integration straightforward.
• The nature of \( R \) decides which coordinate system (Cartesian or polar) might be more convenient for integration.

Understanding your region of integration is essential to accurately setting up and computing a double integral, ensuring you're integrating the intended section of the function.