Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-6, evaluate the iterated integrals. $$ \int_{0}^{2 \pi} \int_{0}^{\theta} r d r d \theta $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{4\pi^3}{3} \).

Step 1: Understand the Integral Setup

The integral given is an iterated integral of the function in polar coordinates. The order of integration is first with respect to \( r \) and then with respect to \( \theta \). The bounds for \( r \) are from 0 to \( \theta \) while the bounds for \( \theta \) are from 0 to \( 2\pi \). This describes a region where the limit for each \( r \) depends on \( \theta \).

Step 2: Integrate with Respect to r

First, integrate the innermost integral \( \int_{0}^{\theta} r\, dr \). Using the power rule for integration, \( \int r\, dr = \frac{r^2}{2} \), we calculate the integral:\[ \int_{0}^{\theta} r\, dr = \left[ \frac{r^2}{2} \right]_{0}^{\theta} = \frac{\theta^2}{2}. \]

Step 3: Integrate with Respect to θ

Next, with the result from Step 2, integrate with respect to \( \theta \):\[ \int_{0}^{2\pi} \frac{\theta^2}{2} \, d\theta. \] The term \( \frac{1}{2} \) is a constant, so the integral simplifies to\[ \frac{1}{2} \int_{0}^{2\pi} \theta^2 \, d\theta. \]

Step 4: Compute the Final Integral

Find the integral \( \int \theta^2 \, d\theta \) using the formula \( \int \theta^n \, d\theta = \frac{\theta^{n+1}}{n+1} \) for \( n=2 \). Compute:\[ \int \theta^2 \, d\theta = \frac{\theta^3}{3}. \]Evaluate the definite integral:\[ \frac{1}{2} \left[ \frac{\theta^3}{3} \right]_{0}^{2\pi} = \frac{1}{2} \left( \frac{(2\pi)^3}{3} - 0 \right) \].

Step 5: Compute the Final Result

Calculate the numeric answer from the expression:\[ \frac{1}{2} \cdot \frac{8\pi^3}{3} = \frac{4\pi^3}{3}. \]

Step 6: Conclusion

The value of the iterated integral is \( \frac{4\pi^3}{3} \).

Key Concepts

Polar Coordinates

Polar coordinates offer a different way to describe locations in a plane based on distance from a fixed point called the origin and an angle from a fixed direction. This is in contrast to Cartesian coordinates, which use x and y axes. In polar coordinates, each point is represented as \( (r, \theta) \) where

• \( r \) is the radial distance from the origin.
• \( \theta \) is the angular distance measured from a reference direction, usually the positive x-axis.

Expressing problems in polar coordinates is especially useful when dealing with circular or radial symmetry. In the original exercise, the problem is set in polar coordinates because the area being integrated has circular bounds, extending from \( 0 \) to \( 2\pi \) for \( \theta \). By employing polar coordinates, the problem becomes more manageable and aligned with the symmetry of the region.

Integration Techniques

When solving iterated integrals, especially those given in polar coordinates, choosing the right integration technique is crucial. Two main considerations are:

• **Order of integration:** Deciding whether to integrate with respect to \( r \) or \( \theta \) first can affect the complexity of the problem. In many cases, the bounds will dictate this choice.
• **Appropriate integration rules:** For instance, using the power rule to integrate \( r \) helps simplify the innermost integral.

In our exercise, the integral \( \int_{0}^{\theta} r \, dr \) was simplified using the power rule, leading to \( \frac{r^2}{2} \). This simplification allowed the problem to move smoothly to the next step. Mastering these techniques is important as it provides a straightforward pathway for evaluating more complex integrals efficiently.

Definite Integrals

Definite integrals compute the accumulation of quantities over a specific interval. In calculus, they essentially measure the "total quantity" that accumulates as we move across the designated region of integration. This is different from indefinite integrals, which provide a general form without fixed boundaries.
In iterated integrals, such as in our example, we perform this process twice:

• First, over the interval for \( r \) (from 0 to \( \theta \)).
• Second, over the interval for \( \theta \) (from 0 to \( 2\pi \)).

The result of a definite integral provides the solution as a specific numerical value, representing the accumulated measure over the entire region. This numeric result is rational and can be applied in practical scenarios to derive insights from the data bounded by those intervals, like the area enclosed within those bounds.

Calculus Problems

Calculus problems often involve applying fundamental concepts to achieve complex results. An iterated integral, like the one in our exercise, challenges students to blend multiple skills:

• Understanding coordinate systems (like polar coordinates) which can simplify integration.
• Accurately applying integration techniques such as changing order or using specific integration rules.
• Evaluating definite integrals to pinpoint precise numerical answers.

In the given problem, each step is a mini-challenge by itself. Yet, through practiced application of calculus principles, the task teaches essential skills:

• The ability to break down multi-element problems into manageable parts.
• The skill to interpret mathematical notation within real-world contexts.
• The power in converting seemingly complex setups into streamlined solutions.

These principles go beyond individual exercises, offering a toolkit for tackling diverse questions in student assignments or real-world scenarios.