Problem 28 - Calculus Early Transcendentals

Question

Sketch the given region of integration \(R\) and evaluate the integral over \(R\) using polar coordinates. $$\iint_{R} e^{-x^{2}-y^{2}} d A ; R=\left\\{(x, y): x^{2}+y^{2} \leq 9\right\\}$$

Step-by-Step Solution

Answer

Question: Sketch the region R and evaluate the integral over the region using polar coordinates, given that R is defined by the inequality \(x^2 + y^2 \leq 9\). Answer: The region R is a circle centered at the origin with a radius of \(3\). The integral over this region is equal to \(\pi - \pi e^{-9}\).

Step 1: Sketch the region R

The given region is defined by the inequality \(x^2 + y^2 \leq 9\). This inequality represents the interior (including the border) of a circle centered at \((0, 0)\) with a radius of \(3\). So, we need to sketch a circle centered at the origin with a radius of \(3\).

Step 2: Convert the given integral to polar coordinates

In order to convert the given integral to polar coordinates, we will use the substitution $$ x = r \cos\theta $$ $$ y = r \sin\theta $$ Also, recall that \(dA = r\, dr\, d\theta\). Given that the region \(R\) is the entire circle, we know that \(r\) will go from \(0\) to \(3\), and \(\theta\) will go from \(0\) to \(2\pi\). Now we can rewrite the integral as $$ \iint_{R} e^{-(r\cos\theta)^{2}-(r\sin\theta)^{2}} r\,d r\, d\theta $$ Simplifying the exponential term, we have $$ \iint_{R} e^{-r^2(\cos^2\theta+\sin^2\theta)} r\,d r\, d\theta $$ Since \(\cos^2\theta+\sin^2\theta=1\), we can further simplify the integral to $$ \int_{0}^{2 \pi} \int_{0}^{3} e^{-r^2} r\,d r\, d\theta $$

Step 3: Evaluate the integral using polar coordinates

First, we will integrate with respect to \(r\): $$ \int_{0}^{3} e^{-r^2} r\,d r = -\frac{1}{2} \int_{0}^{3} e^{-r^2} (-2 r d r) = -\frac{1}{2} \left[ e^{-r^2} \right]_{0}^{3} $$ Evaluating the integral, we get $$ -\frac{1}{2} (e^{-9} - e^{0}) = \frac{1}{2}(1 - e^{-9}) $$ Now, we will integrate with respect to \(\theta\): $$ \int_{0}^{2 \pi} \frac{1}{2}(1 - e^{-9}) d\theta = \frac{1}{2}(1 - e^{-9}) \int_{0}^{2 \pi} d\theta $$ Evaluating the integral, we get $$ \frac{1}{2}(1 - e^{-9})(\theta) \Bigg|_{0}^{2\pi} = (1 - e^{-9}) \pi = \pi - \pi e^{-9} $$

Step 4: Present the final result

The result of the integral over the region \(R\) is $$ \iint_{R} e^{-x^{2}-y^{2}} d A = \pi - \pi e^{-9} $$

Key Concepts

Double Integral

In the realm of calculus, a double integral is a way to integrate over a two-dimensional area. It gives us the volume under a surface where the base of the solid is the region we're integrating over, and the height is given by a function.

In the exercise, we are tasked with finding the integral of the function \(e^{-x^2-y^2}\) over a specified region \(R\). The process involves two integrations: one with respect to \(x\) (or \(r\) in polar coordinates), and another with respect to \(y\) (or \(\theta\) in polar coordinates). These sequential integrations are often represented by nested integral signs, indicating that you perform the integration for one variable while treating the other as a constant, and then integrate the resulting expression with respect to the second variable.

Polar Coordinate System

The polar coordinate system provides an alternative way of representing points in a plane. Instead of using horizontal and vertical coordinates (\(x\) and \(y\)), points are located based on their distance from the origin (\(r\)) and the angle (\(\theta\)) they make with the positive \(x\)-axis.

In the given problem, we convert the function to polar coordinates by replacing \(x\) with \(r\cos(\theta)\) and \(y\) with \(r\sin(\theta)\). This is particularly useful when the region \(R\) is a circle or part of a circle as it simplifies the integration process.

Region of Integration

Every integral is computed over some region, which is called the region of integration. In Cartesian coordinates, this might be defined by inequalities involving \(x\) and \(y\). But for circular and radial regions, polar coordinates often simplify the problem.

In our exercise, the region \(R\) is defined by \(x^2 + y^2 \leq 9\). This translates to the area inside a circle with radius \(3\) in polar coordinates. It simplifies the boundaries for \(r\) and \(\theta\) to \(0 \leq r \leq 3\) and \(0 \leq \theta \leq 2\pi\), respectively.

Area Element in Polar Coordinates

When calculating double integrals, we use an area element to represent a small piece of the region. In Cartesian coordinates, this area element is often written as \(dA = dx\,dy\). However, in polar coordinates, the area element is \(dA = r\,dr\,d\theta\), which accounts for the fact that the same arc length sweeps out a larger area as you move further from the origin.

The exercise demonstrates the importance of multiplying by \(r\) in the polar form of the area element. Without it, we would not be accurately representing the area of each infinitesimal piece of the region.