Question

In Exercises \(15-16,\) special double integrals are presented that are especially well suited for evaluation in polar coordinates. Consider \(\iint_{f} e^{-\left(x^{2}+y^{2}\right)} d A .\) (a) Why is this integral difficult to evaluate in rectangular coordinates, regardless of the region \(R ?\) (b) Let \(R\) be the region bounded by the circle of radius \(a\) centered at the origin. Evaluate the double integral using polar coordinates. (c) Take the limit of your answer from \((b),\) as \(a \rightarrow \infty\). What does this imply about the volume under the surface of \(e^{-\left(x^{2}+y^{2}\right)}\) over the entire \(x-y\) plane?

Short answer

The volume under the surface is \( \pi \) over the entire \( x-y \) plane.

Step-by-step solution

Analyze the Problem in Rectangular Coordinates

The integral \( \iint_{f} e^{-(x^2+y^2)} \, dA \) is complicated in rectangular coordinates due to the form \( e^{-(x^2+y^2)} \). This expression does not simplify well under Cartesian transformations, making integration challenging without substitution or transformations.

Convert to Polar Coordinates

In polar coordinates, the expression \( x^2 + y^2 \) becomes \( r^2 \), where \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). The Cartesian differential \( dx \, dy \) is \( r \, dr \, d\theta \) in polar coordinates. Thus, the integral becomes \( \iint e^{-r^2} \, r \, dr \, d\theta \).

Set Up the Integral over the Circular Region

The region \( R \) is a circle of radius \( a \), so \( r \) ranges from \( 0 \) to \( a \) and \( \theta \) from \( 0 \) to \( 2\pi \). Therefore, the integral is expressed as \[ \int_{0}^{2\pi} \int_{0}^{a} e^{-r^2} r \, dr \, d\theta. \]

Evaluate the Inner Integral

Evaluate \( \int_{0}^{a} e^{-r^2} r \, dr \) by substituting \( u = r^2 \), so \( du = 2r \, dr \). The limits change accordingly to \( u = 0 \) and \( u = a^2 \). Thus, the integral becomes \( \frac{1}{2} \int_{0}^{a^2} e^{-u} \, du = \frac{1}{2}\left[-e^{-u}\right]_{0}^{a^2} = \frac{1}{2}(1 - e^{-a^2}) \).

Evaluate the Outer Integral

Integrate with respect to \( \theta \): \[ \int_{0}^{2\pi} \frac{1}{2} (1 - e^{-a^2}) \, d\theta = \frac{1}{2} (1 - e^{-a^2}) \cdot 2\pi \]. This simplifies to \( \pi (1 - e^{-a^2}) \).

Take the Limit as \( a \rightarrow \infty \)

Taking \( a \rightarrow \infty \), the expression \( e^{-a^2} \rightarrow 0 \). Therefore, \( \pi (1 - e^{-a^2}) \rightarrow \pi \). This indicates that the integral over the entire \( x-y \) plane equals \( \pi \).

Key concepts

Rectangular Coordinates

In the world of mathematics, rectangular coordinates are commonly known as Cartesian coordinates, which involve plotting points on the two-dimensional plane using the x-axis and y-axis. They are incredibly useful for many applications; however, when it comes to integrating certain functions, they can pose some challenges.

Consider the expression \( e^{-(x^2 + y^2)} \) used in our integral. This form is complex to deal with directly using rectangular coordinates. This complexity arises because the exponent involves a sum of squares, which is not easily manipulated or simplified.

While various integration techniques exist, such as substitution, they may not always simplify the equation effectively. In many cases, converting to a more suitable coordinate system is the key to successful integration.

Polar Coordinates Transformation

Transforming equations from rectangular coordinates to polar coordinates can sometimes greatly simplify the problem. In polar coordinates, every point in the plane is determined by a distance from the origin \( r \), and an angle \( \theta \) from the positive x-axis.

The transformation rules are:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)
• \( x^2 + y^2 = r^2 \)
• The differential area element \( dx \, dy \) becomes \( r \, dr \, d\theta \)

Using these transformations, the integration becomes more manageable. The expression \( e^{-(x^2 + y^2)} \) simplifies to \( e^{-r^2} \). This results in an integral over a simpler form, \( \iint e^{-r^2} r \, dr \, d\theta \), facilitating the integration process.

Exponential Function Integration

Handling the integral involving an exponential function in polar coordinates is straightforward due to the simplifications we achieved through the transformation. The process involves first working on the inner integral, \( \int e^{-r^2} r \, dr \).

By substituting \( u = r^2 \), which gives us \( du = 2r \, dr \), we can change the variable of integration. The limits of integration adjust accordingly from \( r = 0 \) to \( r = a \), becoming \( u = 0 \) to \( u = a^2 \). The integral then becomes \( \frac{1}{2} \int e^{-u} \, du \).

Evaluating this gives \( \frac{1}{2} \left[ -e^{-u} \right]_{0}^{a^2} = \frac{1}{2}(1 - e^{-a^2}) \). The cleaner form, \( 1 - e^{-a^2} \), helps us seamlessly proceed to evaluating the outer integral for the complete solution.

Limit Evaluation

Taking limits is a crucial step in calculus, especially when evaluating integrals over unbounded regions. In the context of our problem, after evaluating the inner integral, we proceed to calculate the outer integral: \( \int_{0}^{2\pi} \frac{1}{2} (1 - e^{-a^2}) \, d\theta \).

This resolves to \( \frac{1}{2} (1 - e^{-a^2}) \cdot 2\pi = \pi (1 - e^{-a^2}) \). At this juncture of the problem, we are tasked with taking the limit of the integral result as \( a \rightarrow \infty \).

As \( a \to \infty \), \( e^{-a^2} \to 0 \), simplifying our expression to \( \pi (1 - 0) = \pi \). This indicates that the integral over the entire plane converges to \( \pi \), highlighting the extent of the volume under the surface.

Region of Integration

Defining a region of integration is vital to solving any double integral, as it sets the bounds for the variables involved. In this exercise, the region is defined by a circle of radius \( a \) centered at the origin.

In polar coordinates, this setup translates to having \( r \) varying from \( 0 \) to \( a \) and \( \theta \) stretching from \( 0 \) to \( 2\pi \). This ensures that we are integrating over the circular region defined in the problem, maintaining the bounds within which we calculate our values.

Understanding this region is crucial for correctly applying the transformation and solving the integral. It sets the groundwork not only for the integration process but also for seamlessly transitioning into evaluating the limit as \( a \rightarrow \infty \), as discussed earlier.