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 \(\iint f(x, y) d A \geq 0\)

Step-by-Step Solution

Answer

If \(f(x, y) \geq 0\) on \(R\), then \(\iint f(x, y) dA \geq 0\) by the comparison property.

Step 1: Understand the Comparison Property

The comparison property of double integrals states that if a function \(f(x, y)\) is greater than or equal to another function \(g(x, y)\) over a region \(R\), then the double integral of \(f\) over \(R\) is greater than or equal to the double integral of \(g\) over \(R\). Mathematically, this is expressed as: \[ \iint_R f(x, y) \, dA \geq \iint_R g(x, y) \, dA \] if \(f(x, y) \geq g(x, y)\) for all \((x, y) \in R\).

Step 2: Consider the Function g(x,y) = 0

To apply the comparison property, set \(g(x, y) = 0\). Since by assumption \(f(x, y) \geq 0\) over the region \(R\), we have \(f(x, y) \geq g(x, y)\).

Step 3: Apply the Comparison Property

Now substitute \(g(x, y) = 0\) into the comparison property: \[ \iint_R f(x, y) \, dA \geq \iint_R 0 \, dA \] The right-hand side integral, \(\iint_R 0 \, dA\), evaluates to 0 since the integral of zero over any region is zero.

Step 4: Conclude the Inequality

As per the comparison property, we derived \(\iint_R f(x, y) \, dA \geq 0\). Therefore, the integral of \(f(x, y)\) over region \(R\) is greater than or equal to zero, confirming the result when \(f(x, y) \geq 0\).

Key Concepts

Double Integrals

In calculus, a double integral is an integral that is evaluated over a two-dimensional area. More specifically, you are calculating the volume under a surface by integrating over a region defined in two dimensions. The notation for a double integral looks like \iint\ which indicates we're working in multiple dimensions rather than just a line.
A good way to imagine a double integral is to picture a topographical map of a hill. By calculating the double integral, we determine the total volume of space beneath the hill and above a certain plane.

• The process involves integrating a function of two variables, \(f(x, y)\), over a specified two-dimensional region \(R\).
• This allows us to find not just areas and volumes but also the total value summed over a distributed area.

Typically, you apply limits of integration based on the boundaries of the region \(R\). This might be a simple rectangle, or it could be more complex, with curved sides.

Comparison Property

The comparison property of integrals is a helpful tool in the world of calculus that aids in establishing inequalities. This property gives us a way to relate the integrals of two different functions over the same region. Essentially, if you know that one function is always larger than another function across an entire region, then the area 'under' that function is going to be larger too.
How the math looks:

• You have two functions \(f(x, y)\) and \(g(x, y)\).
• If, everywhere on a region \(R\), \(f(x, y) \geq g(x, y)\), then their integrals bear the same relationship: \iint_R f(x, y) \, dA \geq \iint_R g(x, y) \, dA\.

This is a valuable property because it allows you to make conclusions about one integral based on another that you might already know something about.
Applying this can show, for example, that a function greater than or equal to zero will always have a non-negative integral, reaffirming basic ideas of area being non-negative.

Inequalities in Integrals

Understanding inequalities in integrals is crucial when working with double integrals and the comparison property. Realize that when a function is non-negative over a region, its integral also turns out to be non-negative.
Inequalities come naturally here because integrals represent sums (or total areas). So, if \( f(x,y) \geq 0\), meaning the surface doesn't dip below the plane, then the integral gives you the total of non-negative values, which can't come out to be less than zero.

• This is why setting \( g(x, y) = 0 \) in the comparison property illustrates that \(\iint_R f(x, y) \, dA \geq 0\).
• The beauty of using inequalities in this manner is that you don't have to calculate the exact value; you just know what's greater than what.

By showing \( \iint_R f(x, y) \, dA \geq 0 \), we confirm that all the quantities being considered (assuming \( f(x,y) \geq 0\) are essentially 'piling' up above the plane, thus positively contributing to the integral's value.