Problem 30 - University Calculus: Early Transcendentals

Question

Suppose \(f(x, y)\) is continuous and nonnegative over a region \(R\) in the plane with a defined area \(A(R) .\) If \(\iint_{R} f(x, y) d A=0,\) prove that \(f(x, y)=0\) at every point \((x, y) \in R\)

Step-by-Step Solution

Answer

If \( \iint_{R} f(x, y) \ dA = 0 \) and \( f(x, y) \geq 0 \), then \( f(x, y) = 0 \) at all points in \( R \).

Step 1: Understanding the Problem

We are given a continuous, nonnegative function \( f(x, y) \) over a region \( R \) with a defined area \( A(R) \), and its double integral over \( R \) equals zero. We need to prove that \( f(x, y) = 0 \) for all points in \( R \).

Step 2: Analyzing Conditions

Since \( f(x, y) \) is nonnegative, \( f(x, y) \geq 0 \) for all \( (x, y) \in R \). The integral \( \iint_{R} f(x, y) \, dA = 0 \) suggests that the sum of values (the "volume" under \( f(x, y) \)) is zero.

Step 3: Using the Integral Property

The integral \( \iint_{R} f(x, y) \, dA = 0 \) implies that the area under the curve defined by \( f(x, y) \) over \( R \) must also be zero. Since \( f(x, y) \) is nonnegative, in order for the integral to equal zero, \( f(x, y) \) must be zero at every point in \( R \).

Step 4: Applying Continuity

The continuity of \( f(x, y) \) means there are no breaks or jumps in the function over \( R \). Therefore, if \( f(x, y) = 0 \) at any subset of \( R \), by continuity and the integral equaling zero, \( f(x, y) \) must equal zero at all points of \( R \).

Step 5: Conclusion

Since \( f(x, y) \) is both continuous and nonnegative over \( R \), and the double integral evaluates to zero, we conclude that \( f(x, y) = 0 \) at every point \( (x, y) \in R \).

Key Concepts

Continuity

In mathematics, the concept of continuity for a function is fundamental. When we say a function is continuous over a region in the plane, it means that the function does not have any breaks, jumps, or abrupt changes within that region. This property is akin to drawing a line without lifting your pen from the paper.

In the context of our problem, we have a function \( f(x, y) \) that is continuous over a region \( R \) in the plane. This continuity ensures that if \( f(x, y) \) equals zero at any point and the double integral of the function over the entire region is zero, then \( f(x, y) \) must be zero throughout \( R \).

• If \( f(x, y) \) had a discontinuity, it could "jump" from zero to a positive value somewhere in \( R \), potentially affecting the integral's result.
• Since there are no such jumps, the function must be consistently zero everywhere under these settings.

Nonnegative Function

A nonnegative function is one that never takes on negative values for any inputs within its domain. More simply, the function never "dips below" the horizontal axis on a graph. For \( f(x, y) \), being nonnegative over region \( R \) means \( f(x, y) \geq 0 \) for all points \( (x, y) \) in \( R \).

This attribute is crucial in the given problem because it ensures that all integral contributions are either zero or positive. Thus, a nonnegative function can only yield an integral result of zero if it is zero everywhere within the region.

• Since every tiny piece of \( R \) is contributing \( f(x, y) \cdot dA \geq 0 \) to the integral, the only possibility for the sum to equal zero is if every piece is zero.

Region in the Plane

A region in the plane, denoted as \( R \), is simply a specific portion of the two-dimensional plane under consideration. It can have a defined area \( A(R) \) and often has boundaries that enclose it.

In our problem, \( R \) is the area over which the double integral is performed, and it's assumed to have a finite area. The double integral \( \iint_{R} f(x, y) \, dA \) calculates what we can think of as the "volume" under the surface defined by \( f(x, y) \) over this region.

• The state of \( R \) as a region with an area adds context to why the integral might yield zero if \( f(x, y) \) is always zero across it.
• If the function were zero only at discrete points or within small subregions, the broader continuity and nonnegative properties across \( R \) solidify the function being zero everywhere.