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. $$\iint_{R} e^{-x^{2}-y^{2}} d A ; R=\\{(r, \theta): 0 \leq r < \infty, 0 \leq \theta \leq \pi / 2\\}$$

Short answer

Question: Evaluate the integral \(\iint_{R} e^{-x^{2}-y^{2}} dA\) over the region \(R \subset \mathbb{R}^{2}\), where \(R\) is the quarter disk defined by the polar coordinate conditions \(0 \leq r < \infty\) and \(0 \leq \theta \leq \pi/2\). Answer: \(\frac{\pi}{4}\)

Step-by-step solution

Convert the given function from Cartesian to Polar coordinates

First, remember the conversion relationship between Cartesian coordinates \((x, y)\) and Polar coordinates \((r, \theta)\): $$ x = r \cos{\theta} $$ $$ y = r \sin{\theta} $$ Now convert the function \(f(x, y) = e^{-x^2 - y^2}\) to polar coordinates by substituting \(x = r \cos{\theta}\) and \(y = r \sin{\theta}\): $$ f(r, \theta) = e^{-r^2(\cos^2{\theta}+\sin^2{\theta})} $$ Since \(\cos^2{\theta} + \sin^2{\theta} = 1\), we have: $$ f(r, \theta) = e^{-r^2} $$

Substitute the given limits into the integral and evaluate the integral

Now, substitute the limits of the region \(R\) into the integral, which is given by: $$ \iint_{R} e^{-x^{2}-y^{2}} dA = \int_{0}^{\pi/2} \int_{0}^{\infty} f(r, \theta) r dr d\theta = \lim_{b \to \infty} \int_{0}^{\pi/2} \int_{0}^{b} e^{-r^2} r dr d\theta $$ Now, we can evaluate the integral step by step: $$ \lim_{b\to\infty} \int_{0}^{\pi/2}\int_{0}^{b} e^{-r^2}r d r d \theta = \lim_{b\to\infty}\int_{0}^{\pi/2}\left[-\frac{1}{2} e^{-r^2}\right]_{0}^{b} d\theta $$ Now evaluate the limits of the radial integral: $$= \lim_{b\to\infty} \int_{0}^{\pi/2}\left(-\frac{1}{2} e^{-b^2} + \frac{1}{2} e^{0}\right)d\theta = \lim_{b\to\infty} \int_{0}^{\pi/2}\left(-\frac{1}{2} e^{-b^2} + \frac{1}{2}\right)d\theta$$ Evaluate the θ-integral: $$= \lim_{b\to\infty}\left[-\frac{1}{2} e^{-b^2}\theta + \frac{1}{2}\theta\right]_{0}^{\pi/2} = \lim_{b\to\infty}\left(\frac{\pi}{4} - \frac{1}{2}e^{-b^2}\frac{\pi}{2}\right)$$ Since the limit of \(e^{-b^2}\) as \(b \to \infty\) is 0: $$ = \frac{\pi}{4} - \frac{1}{2}\cdot 0 \cdot \frac{\pi}{2} = \frac{\pi}{4}$$ So, the value of the integral is \(\boxed{\frac{\pi}{4}}\).

Key concepts

Polar Coordinate System

Understanding the polar coordinate system is fundamental when dealing with problems in a plane that involve circular or rotational symmetry. Unlike the rectangular Cartesian coordinate system which uses x and y axes for positioning, the polar coordinate system identifies points based on their distance from a reference point (the origin) and the angle from a reference direction, usually the positive x-axis.

In this system, any point P can be represented as \(r, \theta\), where \(r\) is the radial distance from the origin to the point, and \(\theta\) is the angle measured in radians from the reference direction to the line segment joining the point to the origin. The polar coordinate system is especially useful for simplifying complex integrals and for analyzing figures with radial symmetry, which are common in applications such as physics, engineering, and navigation.

Double Integrals

A double integral allows us to compute the volume under a surface over a region in the xy-plane. Conceptually, we can think of a double integral as a way to add up small 'slices' of volume over a particular region. The double integral over a region R for a function f(x, y) is denoted as \( \iint_R f(x, y) \, dA \), where \( dA \) is a small area element within the region R.

When we convert to polar coordinates, our double integral includes an extra factor of \( r \) because \( dA = r \, dr \, d\theta \). This factor of \( r \) compensates for the fact that areas further from the origin in polar coordinates are larger than those closer to it. As a result, evaluating double integrals in polar coordinates is particularly efficient for regions that are circles, sectors of circles, or annuli.

Limits of Integration

The limits of integration define the region over which we integrate. In a double integral, these limits represent the bounds of the region R in the xy-plane. In Cartesian coordinates, the limits are typically constants or functions of x and y. However, in polar coordinates, the radial limits can range from a constant value to infinity, which requires the use of improper integrals.

When we deal with improper integrals in polar coordinates, often we have a limit where \( r \) approaches infinity, representing an unbounded region. To evaluate these integrals, we typically take the limit of an integral as one of the bounds approaches infinity, handling the unboundedness with a limiting process and ensuring the integral converges to a finite value.

Converting Cartesian to Polar Coordinates

Converting from Cartesian to polar coordinates involves using the relationships \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). This transformation is essential for simplifying the integration process for functions involving circular or rotational symmetry.

During the conversion, we substitute these expressions for x and y into the original function, if it’s given in Cartesian form. Also, we replace the area element \( dA \) with \( r dr d\theta \) to reflect the change in the coordinate system. This step often simplifies the function and the bounds of integration, making it easier to evaluate the integral.

Evaluating Improper Integrals

Improper integrals occur when the interval of integration is infinite or when the function approaches infinity within the interval. In the context of polar coordinates, we often encounter integrals where the radial coordinate \( r \) extends to infinity.

To evaluate such integrals, we replace the infinite bound with a variable like \( b \), and then calculate the limit of the resulting proper integral as \( b \) approaches infinity. This process allows us to determine whether the integral converges to a finite number or diverges to infinity. If it converges, we can find the exact value by calculating this limit, as was demonstrated in the exercise problem. Such techniques are invaluable in fields like physics and engineering, where they are used to model phenomena with infinite domains.