Problem 43 - University Calculus: Early Transcendentals

Question

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 over the disk \(x^2+y^2\leq\frac{3}{4}\) is \(\pi \ln(4)\). The integral over \(x^2+y^2\leq 1\) does not exist due to divergence.

Step 1: Recognize the Domain

We need to integrate the function \( f(x, y) = \frac{1}{1 - x^2 - y^2} \) over the disk \( x^2 + y^2 \leq \frac{3}{4} \). This disk is centered at the origin with a radius of \( \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \).

Step 2: Convert to Polar Coordinates

The disk \( x^2 + y^2 \leq \frac{3}{4} \) suggests that polar coordinates will simplify the integral. Let \( x = r \cos \theta \) and \( y = r \sin \theta \). Consequently, the integral becomes:\[\int_0^{2\pi} \int_0^{\frac{\sqrt{3}}{2}} \frac{r}{1 - r^2} \, dr \, d\theta\]

Step 3: Solve the Inner Integral

We focus on the integral \( \int_0^{\frac{\sqrt{3}}{2}} \frac{r}{1 - r^2} \, dr \). Using the substitution \( u = 1 - r^2 \), \( du = -2r \, dr \), we find:\[\int \frac{r}{1 - r^2} \, dr = -\frac{1}{2} \int \frac{1}{u} \, du = -\frac{1}{2} \ln |u|\]Evaluating the limits, we get:\[-\frac{1}{2} \ln \left| 1 - \left(\frac{\sqrt{3}}{2}\right)^2 \right| + \frac{1}{2} \ln |1|\]

Step 4: Evaluate Boundaries

The evaluation simplifies to:\[-\frac{1}{2} \ln \left| 1 - \frac{3}{4} \right| = -\frac{1}{2} \ln \left| \frac{1}{4} \right| = \frac{1}{2}\ln(4)\]Thus, the inner integral evaluates to \( \frac{1}{2}\ln(4) \).

Step 5: Complete the Integration

Now, integrate with respect to \( \theta \):\[\int_0^{2\pi} \left( \frac{1}{2}\ln(4) \right) \, d\theta = \frac{1}{2}\ln(4) \cdot 2\pi = \pi \ln(4)\]This is the value of the integral over the disk \( x^2 + y^2 \leq \frac{3}{4} \).

Step 6: Analyze Integral Over a Larger Disk

For the disk \( x^2 + y^2 \leq 1 \), the function becomes undefined as \( r \rightarrow 1 \), causing the denominator to approach zero. This results in the integral diverging, meaning that the integral does not exist over this larger disk.

Key Concepts

Polar Coordinates

When working with double integrals over circular regions, converting Cartesian coordinates (\(x, y\)) to polar coordinates (\(r, \theta\)) can greatly simplify the process. In polar coordinates, any point in a plane is described by its distance from the origin, \(r\), and the angle, \(\theta\), from the positive \(x\)-axis. The transformation equations are \(x = r \cos \theta\) and \(y = r \sin \theta\). The advantage here is that the circular boundary \(x^2 + y^2 \leq \frac{3}{4}\) changes into a more manageable form, such as \(r \leq \frac{\sqrt{3}}{2}\).

This transformation simplifies not only boundaries but also the integral setup. The area element in Cartesian coordinates, \(dx \, dy\), translates to \(r \, dr \, d\theta\) in polar terms. This means, once variables are swapped, the integral over a disk takes the form:

• \(\int_0^{2\pi} \int_0^{\frac{\sqrt{3}}{2}} \frac{r}{1 - r^2} \, dr \, d\theta\)

This new form makes it easier to handle the integration over the given circular region.

Change of Variables

Change of variables is an essential technique in calculus, especially for evaluating complex integrals. When dealing with integrals in certain regions, transforming the variables can simplify both the integral and the limits. In essence, it involves substituting original variables with newly defined ones that convert an integral into a more doable form.

In our case, using \(u\) to simplify the integral \(\int \frac{r}{1 - r^2} \, dr\) was crucial. By letting \(u = 1 - r^2\), and thus \(du = -2r \, dr\), we made the integration straightforward. The integral then became:

• \(-\frac{1}{2} \int \frac{1}{u} \, du\)

This results in the expression \(-\frac{1}{2} \ln |u|\). Such transformations are vital as they reduce the complexity of evaluating the integral over a set range. Furthermore, they maintain the original problem’s constraints while making the calculations more manageable.

Improper Integrals

Improper integrals often arise when the function to be integrated becomes unbounded or infinite over the region of integration. In particular, for the function \(f(x, y) = \frac{1}{1 - x^2 - y^2}\), it has a problematic point at the boundary where \(x^2 + y^2 = 1\).

For the disk \(x^2 + y^2 \leq 1\), we encounter issues as \(r\) approaches 1, causing the denominator to go towards zero, which in turn leads the function to tend toward infinity. This results in a divergence, meaning that the integral does not converge to a finite number.

• It implies the integral "doesn't exist" over this larger disk.

Understanding the behavior of integrals, both finite and improper, is crucial as it shapes how domains are chosen or modified for solvability. The ability to predict an integral's behavior when nearing critical points determines its effectiveness and validity in applications.