Problem 63 - Calculus Early Transcendentals

Question

Improper integrals arise in polar coordinates when the radial coordinate \(r\) becomes arbitrarily large. Under certain conditions, these integrals are treated in the usual way: $$\int_{\alpha}^{\beta} \int_{a}^{\infty} f(r, \theta) r d r d \theta=\lim _{b \rightarrow \infty} \int_{\alpha}^{\beta} \int_{a}^{b} f(r, \theta) r d r d \theta$$ Use this technique to evaluate the following integrals. $$\int_{0}^{\pi / 2} \int_{1}^{\infty} \frac{\cos \theta}{r^{3}} r d r d \theta$$

Step-by-Step Solution

Answer

Question: Evaluate the improper integral in polar coordinates: $$\int_{0}^{\pi / 2} \int_{1}^{\infty} \frac{\cos \theta}{r^{3}} r d r d \theta$$ Answer: The value of the improper integral is 1.

Step 1: Write the Given Integral

The exercise gives us, $$\int_{0}^{\pi / 2} \int_{1}^{\infty} \frac{\cos \theta}{r^{3}} r d r d \theta$$

Step 2: Separate the Integral into Two Integrals

We can separate the given integral into two parts: one with respect to \(r\) and the other with respect to \(\theta\) as follows: $$\lim _{b \rightarrow \infty} \int_{0}^{\pi / 2}\left(\int_{1}^{b} \frac{\cos \theta}{r^{2}} d r\right) d \theta$$

Step 3: Evaluate the Integral with respect to r

Now, we need to solve the inner integral with respect to \(r\): $$\int_{1}^{b} \frac{\cos \theta}{r^{2}} d r$$ Since \(\cos \theta\) doesn't depend on \(r\), it can be treated as a constant for this integral: $$\cos \theta \int_{1}^{b} \frac{1}{r^{2}} d r$$ Evaluating this integral yields: $$\left[-\frac{\cos \theta}{r}\right]_1^b = -\frac{\cos \theta}{b} + \cos \theta$$ Now, we rewrite the integral in terms of the result: $$\lim _{b \rightarrow \infty} \int_{0}^{\pi / 2}\left(-\frac{\cos \theta}{b} + \cos \theta\right) d \theta$$

Step 4: Evaluate the Integral with respect to θ

Now, we need to evaluate the outer integral with respect to \(\theta\): $$\int_{0}^{\pi / 2}\left(-\frac{\cos \theta}{b} + \cos \theta\right) d \theta$$ We can separate this integral into two parts: $$\lim_{b \rightarrow \infty}\left(-\frac{1}{b}\int_{0}^{\pi / 2}\cos \theta d \theta + \int_{0}^{\pi / 2}\cos \theta d \theta\right)$$ Evaluate each part separately, $$-\frac{1}{b}\left[\sin \theta\right]_0^{\pi / 2} + \left[\sin \theta\right]_0^{\pi / 2} = -\frac{1}{b}(1-0) + (1-0) = 1 - \frac{1}{b}$$ Now, we rewrite the integral in terms of the result: $$\lim _{b \rightarrow \infty}\left(1 -\frac{1}{b} \right)$$

Step 5: Take the Limit as b goes to Infinity

Finally, we need to take the limit as \(b\) goes to infinity: $$\lim _{b \rightarrow \infty}\left(1 -\frac{1}{b} \right)$$ As \(b\) goes to infinity, the fraction \(\frac{1}{b}\) goes to 0, so the limit is: $$1 - 0 = 1$$ Therefore, the improper integral is equal to 1: $$\int_{0}^{\pi / 2} \int_{1}^{\infty} \frac{\cos \theta}{r^{3}} r d r d \theta = 1$$

Key Concepts

Polar Coordinates

Polar coordinates are used to describe the position of a point in a plane using two values: a radial coordinate, denoted as \(r\), and an angular coordinate, \(\theta\). The radial coordinate represents the distance from a fixed center point called the origin, and the angular coordinate measures the angle between a reference direction and the line connecting the point to the origin. This system is useful in situations where circular or spherical symmetry is present, like the problems involving improper integrals as given in the exercise.

The transformation from Cartesian coordinates \((x, y)\) to polar coordinates is given by:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

This conversion makes it easier to integrate functions over circular regions. Specifically, the exercise we solved involves integrating over an infinite radial distance, showcasing an application of polar coordinates in evaluating such an integral.

Integration Techniques

Integration techniques are mathematical methods used to compute integrals, which represent areas under curves or the accumulation of quantities. In our context, one useful technique is the separation of variables to solve multiple integrals. When dealing with the given integral, it's split into a product of an integral in terms of \( r \) and another in terms of \( \theta \).

This is done by expressing the double integral as a nested operation, where the inner integral is evaluated first with respect to \( r \), treating \( \theta \) as a constant, and then the outer integral is evaluated with respect to \( \theta \). Other common techniques include:

• Substitution: changing variables to simplify the integral.
• Partial fraction decomposition: breaking down complicated fractions into simpler parts.
• Integration by parts: used when the integrand is a product of functions.

In the exercise, we used a straightforward separation of variables technique to simplify our calculations and solved the integral efficiently.

Limits

In calculus, limits help us understand the behavior of functions as inputs approach certain values, which is fundamental in evaluating improper integrals. An improper integral is one where either the interval or the function itself goes to infinity. In our problem, the radial coordinate \(r\) extends towards infinity, making the double integral improper.

We addressed this by introducing a limit for the integral from 1 to infinity as follows:\[\lim _{b \rightarrow \infty} \int_{1}^{b} f(r, \theta) \, d r\]By solving the integral over a finite range (1 to \( b \)) first, and then evaluating the limit as \( b \) approaches infinity, we ensure a meaningful result. This approach elegantly handles the undefined behavior that arises when directly evaluating over an infinite range, leading to the conclusion that the improper integral is equal to 1 after applying this limiting process.

Multiple Integrals

Multiple integrals involve evaluating integrals over more than one variable, such as two-dimensional integrals known as double integrals. These integrations are often performed over rectangular regions, or more complex shapes when using coordinate transformations like the polar coordinates.

In our exercise, a multiple integral is needed to evaluate the given function over a polar coordinate system. The concerned double integral is:\[\int_{0}^{\pi / 2} \int_{1}^{\infty} \frac{\cos \theta}{r^{3}} r \, d r \, d \theta\]Here, integration is done first over \( r \), the radial direction, and then over the angular direction \( \theta \). This process is beneficial when dealing with circular or spherical symmetry, allowing us to take advantage of the simplified region of integration that the transformation to polar coordinates provides.

The successful application of multiple integrals in polar coordinates allows us to manage complex geometries and infinite bounds, like the given problem, concluding with a clean and actionable result.