Problem 33 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Sketch the region inside the limaçon given by \(r = 1 + \frac{1}{2} \cos \theta\), and express the double integral over the region \(R\) in terms of an iterated integral. Answer: The region inside the limaçon can be sketched by plotting the polar equation \(r = 1 + \frac{1}{2} \cos \theta\). The double integral of \(f(r, \theta)\) over the region \(R\) can be expressed as an iterated integral in the form: $$ \iint_R f(r, \theta)dA = \int_{\theta=0}^{2\pi} \left( \int_{r=0}^{1+\frac{1}{2}\cos\theta} f(r,\theta)r\,dr\right) d\theta $$

Step 1: Sketch the limaçon

To sketch the limaçon, plot the equation \(r = 1 + \frac{1}{2} \cos \theta\) in polar coordinates. It is important to note that the limaçon is symmetric about the polar axis and has a loop on the right side of the origin.

Step 2: Identify the bounds of the region R

Since we are considering the region inside the limaçon, the bounds of the region \(R\) would be the values at which the limaçon intersects itself. To find this intersection point, set \(r = 1 + \frac{1}{2} \cos \theta\) and find the \(\theta\) values where the intersection occurs. The intersection occurs at \(\theta = \pi\), as this is where the loop is formed. The bounds of the region \(R\) are defined as follows: - \(r\) goes from 0 to the limaçon equation \(1 + \frac{1}{2} \cos \theta\). - \(\theta\) varies from 0 to \(2\pi\) which covers the full rotation around the origin.

Step 3: Set up the double integral in polar coordinates

To express the double integral of \(f(r, \theta)\) over region \(R\) in polar coordinates, we substitute \(dA = rdrd\theta\) to the integral. The integral will look like the following: $$ \iint_R f(r, \theta)dA = \int_{\theta=0}^{2\pi} \int_{r=0}^{1+\frac{1}{2}\cos\theta} f(r,\theta)r\,dr\,d\theta $$

Step 4: Write the double integral as an iterated integral

Finally, we need to express the double integral as an iterated integral, of which the general form would be: $$ \iint_R f(r, \theta)dA = \int_{\theta=0}^{2\pi} \left( \int_{r=0}^{1+\frac{1}{2}\cos\theta} f(r,\theta)r\,dr\right) d\theta $$ This expression represents the double integral of the function \(f(r, \theta)\) over the region \(R\), which is the region inside the limaçon \(r = 1 + \frac{1}{2} \cos \theta\).

Key Concepts

Limaçon

A limaçon is a type of polar graph with distinct characteristics based on its equation. In our scenario, the limaçon is represented by the equation \( r = 1 + \frac{1}{2} \cos \theta\). Polar coordinates are ideal for illustrating limaçons due to their circular nature. This particular limaçon is symmetric about the horizontal axis and is notable for the loop it forms.
The term "limaçon" originates from the French word for "snail" due to its characteristic snail-like shape. There are key features to recognize:

• It is symmetric about the polar axis.
• The curve has a loop for certain values of \( \theta \).
• For the given equation, the loop occurs when \( \theta = \pi \).

Plotting such a graph requires evaluating \( r \) at different \( \theta \) values, which demonstrates how the curve changes, forming both the main outer loop and the smaller inner loop. Understanding this shape helps in visualizing different regions within the limaçon for further calculations.

Iterated Integral

Iterated integrals are a method of integrating a function over a specified region, where the integral is performed progressively over one variable at a time. In polar coordinates, this usually means integrating over \( r \) and then \( \theta \), or vice versa.

To set up an iterated integral for a polar region, the bounds must be carefully considered. Typically, they correspond to the region's boundaries.

• The inner integral is with respect to \( r \), going from 0 to the limaçon's boundary equation \( 1 + \frac{1}{2} \cos \theta \).
• The outer integral is with respect to \( \theta \), covering a full rotation range from 0 to \( 2\pi \).

Correctly setting these bounds is crucial for evaluating the integral successfully. Using iterated integrals provides a structured way to approach multi-variable calculus problems, ensuring each dimension is considered step-by-step.

Double Integral

A double integral allows for the evaluation of a function over a two-dimensional region. In polar coordinates, this integral considers both radial and angular components. In simpler terms, it sums up infinitely small regions over the entire area. If our region \( R \) is bounded by a limaçon, we would set up the double integral accordingly.
The double integral is adjusted for polar coordinates by including an additional \( r \) term in the differential area element \( dA \), giving us \( dA = rdrd\theta \).For our limaçon, the double integral is expressed as:\[\iint_R f(r, \theta)dA = \int_{\theta=0}^{2\pi} \left( \int_{r=0}^{1+\frac{1}{2}\cos\theta} f(r,\theta)r\,dr\right) d\theta\]This means we first integrate \( f(r, \theta)r \) with respect to \( r \), considering \( \theta \) as constant, and then with respect to \( \theta \) from 0 to \( 2\pi \).
The concept of double integrals in polar coordinates is powerful for regions like the limaçon, as it simplifies otherwise complex calculations into manageable steps involving radial and angular components.