Problem 7 - College Algebra and Calculus: An Applied Approach

Question

Sketch the region of integration and evaluate the double integral. $$ \int_{-a}^{a} \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2-x^{2}}}} d y d x $$

Step-by-Step Solution

Answer

The region of integration is a circle of radius \( a \) centered at the origin. The double integral evaluates to \( \pi a^2 \).

Step 1: Sketch the Region of Integration

The region of integration corresponds to \( -a \leq x \leq a \) and \( -\sqrt{a^2 - x^2} \leq y \leq \sqrt{a^2 - x^2} \), which are the equations of a circle with radius \( a \) centered at the origin in Cartesian coordinates. So sketch a circle with radius \( a \) centered at the origin.

Step 2: Convert to Polar Coordinates

The double integral becomes easier to solve if we convert to polar coordinates. In polar coordinates, \( x = r\cos{\theta} \) and \( y = r \sin{\theta} \) with \( r^2 = x^2 + y^2 \) and \(-\pi \leq \theta \leq \pi\). The limits of \( r \) will be from 0 to \( a \) and \(\theta\) will be from 0 to \( 2\pi \). The differential area element \( dx \cdot dy \) in polar coordinates becomes \( r \cdot dr \cdot d\theta \). So, the double integral can be rewritten in polar coordinates as \( \int_{0}^{2\pi}\int_{0}^{a} r \cdot dr \cdot d\theta \)

Step 3: Evaluate the Integral

Evaluating the integral, we first perform the \( r \) integral, which gives \( \frac{1}{2}r^2 \). Evaluating this at the limits 0 and \( a \) gives \( \frac{1}{2}a^2 \). We then perform the \( \theta \) integral of \( \frac{1}{2}a^2 \) from 0 to \( 2\pi \) which gives \( \pi a^2 \). Thus, the final result is \( \pi a^2 \).

Key Concepts

Polar Coordinates

Working with double integrals can sometimes be challenging when using Cartesian coordinates, especially if the region of integration is circular. In such cases, polar coordinates offer a convenient alternative.
- Polar coordinates are a way of representing points in a plane using the distance from a reference point (usually the origin) and an angle from a reference direction (typically the positive x-axis).
- This system is especially useful for problems involving circles or curves that emanate from a central point.
In polar coordinates, any point can be described using the pair \(r, \theta\), where \(r\) is the radius or distance from the origin and \(\theta\) is the angle in radians.
For a transformation from Cartesian to polar coordinates:

• The x-coordinate becomes \(x = r\cos(\theta)\).
• The y-coordinate becomes \(y = r\sin(\theta)\).

The area element \(dx \, dy\) in Cartesian coordinates transforms to \(r \, dr \, d\theta\) in polar coordinates, allowing for more straightforward integration in circular regions.

Region of Integration

In the context of double integrals, the region of integration is the area over which the function is being integrated. Identifying this region accurately is crucial for solving the integral.
- Sketching the region of integration helps in visualizing the bounds for integration.
- In the given exercise, the region is described by the limits \(-a \leq x \leq a\) and \(-\sqrt{a^2 - x^2} \leq y \leq \sqrt{a^2 - x^2}\).
This region describes a circle with a radius \(a\) centered at the origin.
- By sketching a circle, one can easily see the symmetry and design the integration limits either using Cartesian or polar coordinates.
In our exercise, the best approach is converting these bounds into polar coordinates, simplifying the integration process as the circular nature of the region fits naturally with polar descriptions.

Cartesian Coordinates

While polar coordinates simplify problems with circular regions, Cartesian coordinates are still crucial for many integrations.
- Cartesian coordinates describe each point using an x-value and a y-value, which is straightforward for rectangular regions.
Spotting the need for a different coordinate system can save considerable time and effort.
- In the original exercise, recognizing the region's circular shape indicates the potential for a polar coordinate approach.
However, beginning with Cartesian coordinates:

• Provides clear structure with simple limits for rectangular regions.
• Is fundamental, as most equations initially exist or are given in this form.

Transformations between Cartesian and polar systems require understanding both coordinates, allowing flexible approaches to a variety of integration problems.