Question

Sketch the following regions \(R\). Then express \(\iint_{R} f(r, \theta) d A\) as an iterated integral over \(R.\) The region outside the circle \(r=\frac{1}{2}\) and inside the cardioid \(r=1+\cos \theta\)

Short answer

Based on the given polar equations, the region R is an area lying between the circle \(r = \frac{1}{2}\) and the cardioid \(r = 1 + \cos \theta\). After sketching the region, finding the points of intersection, and determining the bounds for integration, we can express the iterated integral over the region R as follows: $$ \iint_R f(r, \theta)dA = \int_{\frac{2}{3}\pi}^{\frac{4}{3}\pi} \int_{\frac{1}{2}}^{1 + \cos \theta} f(r, \theta) rdrd\theta $$

Step-by-step solution

Sketch the Region

To sketch the given region, start by plotting the polar equation \(r = \frac{1}{2}\), which represents a circle centered at the origin with radius \(\frac{1}{2}\). Now, plot the cardioid \(r = 1 + \cos \theta\). Note that the cardioid graph starts from the origin and extends around the circle. The region R lies between these two plots.

Find the Bounds for \(\theta\)

Now we need to find the points where the circle and the cardioid intersects. Set \(r = \frac{1}{2}\) and \(r = 1 + \cos \theta\) equal and solve for \(\theta\): $$ \frac{1}{2} = 1 + \cos \theta $$ $$ \cos \theta = -\frac{1}{2} $$ Thus, \(\theta = \frac{2}{3}\pi\) and \(\theta = \frac{4}{3}\pi\). These two angles are the bounds for the \(\theta\) in the double integral.

Find the Bounds for \(r\)

The bounds for \(r\) will be from the circle's equation to the cardioid's equation in the region R. So, we have \(r\) ranging from \(\frac{1}{2}\) to \(1 + \cos \theta\).

Write the Iterated Integral

Now that we have the bounds for \(r\) and \(\theta\), we can write the iterated integral \(\iint_R f(r, \theta)dA\) as $$ \int_{\frac{2}{3}\pi}^{\frac{4}{3}\pi} \int_{\frac{1}{2}}^{1 + \cos \theta} f(r, \theta) rdrd\theta $$

Key concepts

Double Integral

When we talk about a double integral, we are essentially discussing the integration process over a two-dimensional region. This technique is used to calculate quantities like area, volume, and more complex functions across surfaces.
In this context, the double integral is expressed as \[ \iint_R f(x, y) \, dA \] where * \(R\) represents the region over which we are integrating. * \(f(x, y)\) is the function being integrated over the region \(R\). * \(dA\) indicates the differential area element.
When working with polar coordinates, we express this integral in terms of \(r\) and \(\theta\), as shown in the original exercise. Depending on the region's shape and symmetry, switching to polar coordinates can often simplify calculations. Polar coordinates effectively translate circular or spherical problems into computationally convenient integrals.

Iterated Integral

In mathematics, an iterated integral is a method of repeatedly integrating a function. We perform double integration by evaluating one integral, then using its result as the function for a second integration.
The iterated integral form used in polar coordinates is \[ \int_{\theta_{ ext{min}}}^{\theta_{ ext{max}}} \int_{r_{ ext{min}}}^{r_{ ext{max}}} f(r, \theta) \cdot r \, dr \, d\theta \] This form allows us to calculate the integral over the specified region \(R\) through the bounds defined by \(\theta\) and \(r\).

• The inner integral is the first integral to be computed, usually with respect to \(r\).
• The outer integral computes the result over \(\theta\).

After setting up the bounds based on the given region as shown in the step-by-step solution, we can compute the iterated integral step-by-step. Each partial integration helps narrow down the function to a single value or solution that represents the quantity over the area.

Cardioid

A cardioid is a special kind of curve that looks like a heart shape. The name "cardioid" comes from the Greek word for "heart."
In polar coordinates, a cardioid can be expressed by equations like \(r = 1 + \cos \theta\) or \(r = 1 + \sin \theta\).

• The curve shows symmetry around the polar axis or other specific lines depending on the equation.
• It features a cusp, which is a point where the curve sharply intersects itself and turns outward.

When sketching this curve, it starts from the origin and extends to the maximum radius before curving back. This property makes the cardioid ideal for delimiting regions in polar mathematics, such as the one in the exercise, where it forms the outer boundary of the region \(R\). Understanding its equation and shape is crucial when determined as one of the limits for integration.

Circle in Polar Coordinates

In polar coordinates, circles are one of the basic shapes we can describe. A circle centered at the origin can be represented simply as \(r = a\), where \(a\) is the radius.
Unlike cartesian coordinates, where a circle with radius \(a\) centered at the origin has the equation \(x^2 + y^2 = a^2\), polar coordinates simplify this greatly.
The simplicity of circles in polar coordinates makes them handy for problems involving circular regions, curves, and iterative integration.
In the provided textbook solution, the circle's representation, \(r = \frac{1}{2}\), acts as the inner boundary for the region \(R\) that is being considered in the integral. By understanding these coordinate systems, we can seamlessly switch between describing and calculating such geometrical shapes, especially when working with irregular boundaries like that of a cardioid.