Problem 39 - Thomas Calculus

Question

Existence Integrate the function \(f(x, y)=1 /\left(1-x^{2}-y^{2}\right)\) over the disk \(x^{2}+y^{2} \leq 3 / 4 .\) Does the integral of \(f(x, y)\) over the disk \(x^{2}+y^{2} \leq 1\) exist? Give reasons for your answer.

Step-by-Step Solution

Answer

The integral exists over the disk \( x^2 + y^2 \leq \frac{3}{4} \), but not over \( x^2 + y^2 \leq 1 \) due to a boundary singularity.

Step 1: Understanding Disk Boundaries

Identify the region of integration. The function \( f(x, y) = \frac{1}{1-x^2-y^2} \) is integrated over two different disks. First, we look at the disk defined by \( x^2 + y^2 \leq \frac{3}{4} \). This is a circular area centered at the origin with radius \( \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \). The second region mentioned is the disk with equation \( x^2 + y^2 \leq 1 \), which has a radius \( 1 \). This understanding helps in examining the behavior of the function near these boundaries.

Step 2: Examine Singularity within Integration Region

The function \( f(x, y) \) could have singularities where the denominator becomes zero: \( 1-x^2-y^2 = 0 \), which simplifies to \( x^2+y^2 = 1 \). This potential singularity is at the boundary of the disk \( x^2 + y^2 \leq 1 \), meaning it’s on the edge of this second disk. Within the disk \( x^2 + y^2 \leq \frac{3}{4} \), there are no points where \( x^2+y^2 = 1 \), thus no singularities exist inside this smaller region.

Step 3: Integrability Over Smaller Disk

Since \( f(x, y) \) is continuous and finite inside the region \( x^2+y^2 \leq \frac{3}{4} \), it is integrable over this disk. We can conclude that for the disk \( x^2+y^2 \leq \frac{3}{4} \), the integral exists because the function has no undefined points or infinite values within this disk.

Step 4: Analyzing Integrability Over Larger Disk

For the disk \( x^2 + y^2 \leq 1 \), the function \( f(x, y) \) approaches infinity as \( x^2 + y^2 \to 1 \). This creates a singularity exactly on the boundary of the integration region, which complicates integrating because it can cause the integral to diverge. Since the function isn’t bounded near the boundary \( x^2 + y^2 = 1 \), it indicates that the integral over this larger disk does not exist.

Key Concepts

Integration

In multivariable calculus, integration is a fundamental concept that extends the idea of summing to functions of several variables, like in our exercise here. We are dealing with a function integrated over a defined region, specifically a disk in a two-dimensional space.

Here, we integrate the function \( f(x, y) = \frac{1}{1-x^2-y^2} \) over a disk.

• Integration process: The area over which we integrate is crucial for deciding if the integral converges to a value or not.
• Convergence: If the function is well-behaved (continuous and finite) throughout the region, the integral exists.

Understanding the behavior of \( f(x, y) \) within the given disk is central to evaluating the integrability.

Singularities

Singularities signify points where the function does not behave normally—like when it becomes undefined or infinite. In our function, the potential singularity is when the denominator goes to zero.
The equation \( 1-x^2-y^2 = 0 \) implies \( x^2 + y^2 = 1 \), indicating the boundary where issues may occur:

• Location: This boundary is where the function \( f(x, y) \) tends to infinity.
• Effect: Singularities affect whether an integral can be computed without diverging to infinity.

In our analysis, the smaller disk \( x^2+y^2 \leq \frac{3}{4} \) does not encompass this boundary, thus no singularities affect its integration.

Regions of Integration

Determining regions of integration is essential as they define the bounds within which our function will be calculated. Here, we deal with two regions, each affecting how we approach the problem:

• Smaller Disk: Defined by \( x^2 + y^2 \leq \frac{3}{4} \). The function is continuous in this region, making the integral straightforward.
• Larger Disk: Defined by \( x^2 + y^2 \leq 1 \). The boundary at \( x^2 + y^2 = 1 \) induces issues due to singularities, challenging our integration attempts.

The choice of region dramatically influences whether the integrals we compute will exist or diverge.

Disk Boundaries

Disk boundaries describe the perimeters of the region of integration. These boundaries help us visualize and constrain where we are evaluating our function. They are pivotal in understanding our exercise:

• The boundary of the smaller disk \( x^2 + y^2 = \frac{3}{4} \) remains within a region where the function is finite and integrable.
• The boundary of the larger disk \( x^2 + y^2 = 1 \) is precisely where singularities occur, causing the function to be undefined at the edge.

These boundaries define the line between what's possible to integrate and where issues may arise.