Problem 34 - 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 leaf of the rose \(r=2 \sin 2 \theta\) in the first quadrant

Step-by-Step Solution

Answer

Based on the given problem, the correct answer is: $$ \iint_{R} f(r, \theta)dA = \int_{0}^{\pi/4} \int_{0}^{2\sin(2\theta)} f(r, \theta)r dr d\theta $$ This iterated integral represents the double integral over the region R inside the leaf of the rose curve \(r=2\sin(2\theta)\) in the first quadrant.

Step 1: Sketch the curve \(r = 2\sin(2\theta)\)

To sketch the curve, we can plot some points. Note that, as the curve is symmetric about the polar axis, we just need to consider the interval \(0\le\theta\le\frac{\pi}{2}\) to sketch the curve in the first quadrant. A table with some values can help: $\begin{array}{c|c} \theta & r \\ \hline 0 & 0 \\ \pi/4 & 2 \\ \pi/2 & 0 \end{array}$ Now with these points, we can draw the curve in the first quadrant. Each petal of the rose will have an endpoint at the origin and will extend to the maximum value of r, which is 2.

Step 2: Identify the region R and find its bounds

We only need to consider the leaf of the rose in the first quadrant. Then, the region R is enclosed by \(r=0\) (the origin) and \(r=2\sin(2\theta)\). We also know that \(0\le\theta\le\frac{\pi}{2}\) to remain in the first quadrant. To find out where the petal starts and ends, set \(r=0\) and solve: \(0=2\sin(2\theta)\) \(\sin(2\theta)=0\) In the first quadrant \(\theta\) is between 0 and \(\frac{\pi}{2}\). We get that \(\theta_1 = 0\) and \(\theta_2 = \frac{\pi}{4}\) when \(r=0\). So, the bounds for \(\theta\) are \(0\le\theta\le\frac{\pi}{4}\). The bounds for r are given by \(0 \le r \le 2\sin(2\theta)\) for this region.

Step 3: Set up the iterated integral for \(\iint_{R} f(r, \theta)dA\)

To express the double integral as an iterated integral, we first need to write the differential area element \(dA\) in polar coordinates. For polar coordinates, the differential area element is given by \(dA = r dr d\theta\). Now, we can write the iterated integral: $$ \iint_{R} f(r, \theta)dA = \int_{0}^{\pi/4} \int_{0}^{2\sin(2\theta)} f(r, \theta)r dr d\theta $$ The iterated integral represents the double integral over the region R inside the leaf of the rose curve \(r=2\sin(2\theta)\) in the first quadrant.

Key Concepts

Double Integrals

Double integrals are a way to integrate over a two-dimensional area. They allow us to calculate the volume under a surface defined above a region in the plane. To understand the application of double integrals, think of them as a sum of infinitely small areas under a curve.

Here's what you should know about double integrals:

• They are written as \(\iint f(x, y) \, dx \, dy\) to represent the summation over a two-dimensional region \( R \).
• When using polar coordinates, they are expressed as \(\iint f(r, \theta) r \, dr \, d\theta\), where \( r \) is the radius and \( \theta \) the angle. The extra \( r \) factor accounts for converting the differential area in Cartesian coordinates to polar.

Double integrals are particularly useful when dealing with circular or radial symmetry in a problem, where polar coordinates can simplify integration.

Iterated Integrals

An iterated integral describes how a double integral can be separated into consecutive (or iterated) integrations. First, you integrate concerning one variable while treating others as constants. Then you integrate the result with respect to the other variable.

Here are key points about iterated integrals:

• Expression: They are typically expressed as \(\int_a^b \int_c^d f(r, \theta) r \, dr \, d\theta\). This notation shows two nested integrals, with the inner integral solved first.
• Order: The order of integration can sometimes be switched, depending on the function and integration limits, as long as the region's boundaries are correctly adjusted.

In polar coordinates, the iterated integrals become easier to handle in regions that are naturally circular or angular. This is why they are valuable in our rose curve example, where we integrate over a radial and angular section of the plane.

Sketching Regions

Sketching the region of integration is crucial before setting up and solving double or iterated integrals. It helps ensure that the limits for integration correctly encompass the desired area.

For the exercise at hand:

• Region Description: The region of interest is inside a leaf of the rose curve given by \( r = 2 \sin(2\theta) \).
• Quadrant Consideration: Only the first quadrant is considered here, with \( 0 \le \theta \le \pi/2 \).

Before integrating, plot key points such as where \( r \) is zero, using calculated points from the sine function such as when \( \theta = 0, \\pi/4, \\pi/2\). This gives a visual idea of the bounds for \( \theta \) and \( r \), with \( r \) extending up to \( 2\sin(2\theta) \).Simplifying the sketch helps confirm the shape and size of the region, leading to more accurate integrations.