Question

Evaluate by using polar coordinates. Sketch the region of integration first. \(\iint_{S} e^{x^{2}+y^{2}} d A\), where \(S\) is the region enclosed by \(x^{2}+y^{2}=4\)

Short answer

The integral evaluates to \(\pi (e^{4} - 1)\).

Step-by-step solution

Understand the Region of Integration

The region of integration is enclosed by the circle given by the equation \(x^2 + y^2 = 4\). This represents a circle centered at the origin with a radius of 2.

Convert to Polar Coordinates

To evaluate the integral using polar coordinates, we express \(x\) and \(y\) in terms of \(r\) and \(\theta\):\[x = r\cos(\theta), \quad y = r\sin(\theta)\] This transforms \(x^2 + y^2 = r^2\). Therefore, our integrand \(e^{x^2+y^2}\) becomes \(e^{r^2}\).

Set Up the Polar Integral

The circle \(x^2 + y^2 = 4\) in polar coordinates is \(r = 2\). Since \(r\) ranges from 0 to 2 and \(\theta\) covers the full circle from 0 to \(2\pi\), the double integral in polar coordinates is:\[\int_{0}^{2\pi} \int_{0}^{2} e^{r^2} r \, dr \, d\theta\]

Integrate with Respect to r

The inner integral is with respect to \(r\), so evaluate:\[\int_{0}^{2} e^{r^2} r \, dr\]Let \(u = r^2\), then \(du = 2r \, dr\) or \(r \, dr = \frac{1}{2} \, du\). The limits change from \(r=0\) to \(u=0\) and \(r=2\) to \(u=4\):\[\frac{1}{2} \int_{0}^{4} e^u \, du = \frac{1}{2} [e^u]_{0}^{4} = \frac{1}{2} (e^{4} - e^{0}) = \frac{1}{2} (e^{4} - 1)\]

Integrate with Respect to θ

Now multiply the result of the \(r\) integral by the integral with respect to \(\theta\):\[\int_{0}^{2\pi} \frac{1}{2} (e^{4} - 1) \, d\theta = \frac{1}{2} (e^{4} - 1) \cdot \theta \bigg|^{2\pi}_{0} = \frac{1}{2} (e^{4} - 1) \cdot 2\pi = \pi (e^{4} - 1)\]

Combine results to solve the integral

Finally, combine the results as calculated:The double integral evaluates to \(\pi (e^{4} - 1)\).

Key concepts

Double Integral

A double integral is a way to integrate over a two-dimensional region, allowing us to calculate the volume under a surface. Imagine you have a thin sheet. Instead of measuring just along a single line (like in one-dimensional integration), in a double integral, we measure over an entire flat region. This sheet can represent anything like temperature distribution or density.
In mathematical terms, a double integral of a function \( f(x, y) \) over a region \( S \) can be denoted as:

• \( \iint_{S} f(x, y) \, dA \)

Here, \( dA \) represents a tiny area element, and \( S \) is the area over which we're integrating.
Double integrals come in handy when you are dealing with functions of two variables. They help describe the entire area under the surface in a given region of the xy-plane.

Region of Integration

The region of integration defines exactly where we apply the double integral. It plays a crucial role because it signifies the limits within which the integration is done.
In the exercise, the region of integration is given by the equation \( x^2 + y^2 = 4 \). This equation describes a circle in the xy-plane:

• Centered at the origin \((0,0)\)
• With a radius of 2 (because \( r^2 = 4 \) implies \( r = 2 \))

Since the shape is a circle, any point \((x, y)\) satisfied by \( x^2 + y^2 \leq 4 \) lies within this region. This means that when we move on to polar coordinates, it will guide how far and in which directions we integrate.

Circle Equation

In cartesian coordinates, the equation \( x^2 + y^2 = a^2 \) describes a circle with center at the origin \((0,0)\) and radius \( a \).
For the given exercise, the circle is represented by \( x^2 + y^2 = 4 \). This equation makes it clear:

• The circle has its center at the origin.
• The radius of the circle is 2, as \( a^2 = 4 \) means \( a = 2 \).

Understanding circle equations is essential, especially in integration, because it often helps us easily shift to polar coordinates.
With this knowledge, we can more easily define the boundaries of our integration region and recognize symmetrical properties that simplify the integration process.

Polar Integral Conversion

Transforming into polar coordinates is a handy method for integrating functions over circular areas. Here's how it works:

• Convert coordinates: \( x = r\cos(\theta), \quad y = r\sin(\theta) \)
• Switch to polar formula: \( x^2 + y^2 = r^2 \)
• Express the area element: In polar coordinates, \( dA \) becomes \( r \, dr \, d\theta \)

This conversion is particularly useful in this exercise, as the original problem involves a circular region. The function \( e^{x^2+y^2} \) becomes \( e^{r^2} \), allowing us to reframe the double integral into a format that leverages the symmetry and simplicity of the circle.
Using polar coordinates, the integration limits become clearer and the computation often becomes more straightforward, especially for circles or radial regions.