Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 iterated integral is \( \frac{4\pi^3}{3} \).

Step 1: Understand the integral limits

The given integral \( \int_{0}^{2 \pi} \int_{0}^{\theta} r \ d r\ d \theta \) is an iterated integral where \( r \) is integrated with limits from 0 to \( \theta \), and then \( \theta \) is integrated from 0 to \( 2\pi \). This integral evaluates the area under the curve by first integrating with respect to \( r \) and then with respect to \( \theta \).

Step 2: Integrate with respect to \( r \)

The inner integral is \( \int_{0}^{\theta} r \ dr \), which represents an integration with respect to \( r \) over the range from 0 to \( \theta \). The antiderivative of \( r \) is \( \frac{1}{2}r^2 \). Evaluating from 0 to \( \theta \), we get: \[ \int_{0}^{\theta} r \ dr = \left[ \frac{1}{2}r^2 \right]_{0}^{\theta} = \frac{1}{2} \theta^2. \]

Step 3: Integrate with respect to \( \theta \)

Now integrate the result \( \int_{0}^{2 \pi} \frac{1}{2} \theta^2 \ d\theta \). The antiderivative of \( \frac{1}{2} \theta^2 \) is \( \frac{1}{6} \theta^3 \). Evaluating this from 0 to \( 2\pi \): \[ \int_{0}^{2\pi} \frac{1}{2} \theta^2 \ d\theta = \left[ \frac{1}{6} \theta^3 \right]_{0}^{2\pi} = \frac{1}{6} (2\pi)^3 - \frac{1}{6} (0)^3 = \frac{1}{6} (8\pi^3). \]

Step 4: Simplify the result

Simplify the expression \( \frac{1}{6} (8\pi^3) \) to find \( \frac{8\pi^3}{6} = \frac{4\pi^3}{3} \). This is the final result of the iterated integral.

Key Concepts

Integration by parts

Integration by parts is a powerful technique for solving more complex integrals, especially those involving products of functions. It's based on the product rule for differentiation and is typically written as:

• \( \int u \ dv = uv - \int v \ du \)

In this scenario, however, the method isn't directly applied to our iterated integral problem. Instead, we're dealing with iterated integrals, which involve evaluating multiple integrals with specific limits. When you have an integral of a product, consider other methods, such as recognizing a pattern or simplifying through substitution. Still, understanding integration by parts can give you insights into handling challenging integrals when applicable.
In summary, although integration by parts isn’t central to solving this iterated integral directly, it remains an invaluable tool in other integration contexts involving products of functions.

Integral limits

The concept of integral limits is essential in determining the region over which you're integrating. Integral limits define the boundaries of the chosen area or volume and are crucial for setting up the integral correctly.
In our case, we have the iterated integral \( \int_{0}^{2 \pi} \int_{0}^{\theta} r \ dr \ d\theta \).
This represents a two-step process:

• The inner integral \( \int_{0}^{\theta} \) signifies that we integrate with respect to \( r \), ranging from 0 to \( \theta \). This usually corresponds to the radius in polar coordinates.
• The outer integral \( \int_{0}^{2 \pi} \) indicates that the final integration is with respect to \( \theta \), which ranges from 0 to \( 2\pi \), effectively sweeping around a full circle in polar coordinates.

Grasping integral limits allows you to visualize the multi-dimensional region you're working with and ensures the correct application of formulas, especially in multivariable calculus.

Area under the curve

The "area under the curve" refers to the computation of the total area bounded by a curve described within specific limits. In calculus, this often involves integration.
For the given iterated integral, calculating the area under the curve involves evaluating\( \int_{0}^{2 \pi} \int_{0}^{\theta} r \ dr \ d\theta \).
This specific computation relates to finding the area within a sector of a circle when using polar coordinates. By first integrating with respect to \( r \) and then with respect to \( \theta \), you sum up infinitesimal areas to determine the total region's area.
Integrating with respect to \( r \) gives a general region, and then \( \theta \) from 0 to \( 2\pi \) effectively "sweeps" this region around, creating a circular section.
This is particularly useful in physical applications where one needs to determine areas and volumes in radial or circular settings.

Antiderivative

Understanding antiderivatives is key to both definite and indefinite integrals. Essentially, an antiderivative is a function whose derivative is the given function.
In the iterated integral \( \int_{0}^{2 \pi} \int_{0}^{\theta} r \ dr \ d\theta \), the step involving antiderivatives occurs during both integrals:

• First, for \( \int_{0}^{\theta} r \ dr \), the antiderivative of \( r \) is \( \frac{1}{2}r^2 \). Calculating this from 0 to \( \theta \) gives \( \frac{1}{2}\theta^2 \).
• Next, for \( \int_{0}^{2 \pi} \frac{1}{2}\theta^2 \ d\theta \), the antiderivative is \( \frac{1}{6}\theta^3 \). Evaluating this from 0 to \( 2\pi \) results in \( \frac{1}{6}(8\pi^3) \).

Gaining proficiency with antiderivatives allows you to identify how changes in one variable affect an entire system, supporting effective problem-solving in calculus.