Problem 31 - University Calculus: Early Transcendentals

Question

Each gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral. $$\int_{-\pi / 3}^{\pi / 3} \int_{0}^{\sec t} 3 \cos t d u d t \quad(\text { the } t u \text { -plane })$$

Step-by-Step Solution

Answer

The integral evaluates to \(2\pi\).

Step 1: Understand the Integral Limits and Function

The given double integral is \( \int_{-\pi / 3}^{\pi / 3} \int_{0}^{\sec t} 3 \cos t \, du \, dt \). The outer integral \( \int_{-\pi / 3}^{\pi / 3} \) represents the limits for \( t \), while the inner integral \( \int_{0}^{\sec t} \) represents the limits for \( u \). The function to integrate is \( 3 \cos t \). Specifically, for each \( t \), \( u \) varies from \( 0 \) to \( \sec t \).

Step 2: Sketch the Region of Integration

Draw the \( tu \)-plane where the region of integration is bounded. The range for \( t \) is from \(-\pi/3\) to \(\pi/3\). For a given \( t \), \( u \) ranges from 0 to \( \sec t \). The line \( u = \sec t \) is plotted over the interval of \( t = [-\pi/3, \pi/3] \). This creates a region above the line \( u = 0 \) and below the curve \( u = \sec t \).

Step 3: Evaluate the Inner Integral

Evaluate the inner integral \( \int_{0}^{\sec t} 3 \cos t \, du \). Since \( 3 \cos t \) is constant with respect to \( u \), the integral is straightforward:\[3 \cos t \int_{0}^{\sec t} 1 \, du = 3 \cos t [u]_{0}^{\sec t} = 3 \cos t (\sec t - 0) = 3 \cos t \sec t.\]Recall that \( \sec t = \frac{1}{\cos t} \), thus reducing the expression to \( 3 \).

Step 4: Evaluate the Outer Integral

Now evaluate the outer integral:\[\int_{-\pi / 3}^{\pi / 3} 3 \, dt.\]This simplifies to evaluating the definite integral of a constant:\[3 \int_{-\pi / 3}^{\pi / 3} 1 \, dt = 3 \left[t\right]_{-\pi / 3}^{\pi / 3} = 3 \left( \frac{\pi}{3} - \left(-\frac{\pi}{3}\right) \right) = 3 \left( \frac{2\pi}{3} \right) = 2\pi.\]

Step 5: Final Solution

The value of the double integral \( \int_{-\pi / 3}^{\pi / 3} \int_{0}^{\sec t} 3 \cos t \, du \, dt \) is \( 2\pi \). This completes the evaluation of the integral over the specified region in the \( tu \)-plane.

Key Concepts

Understanding the Cartesian Coordinate Plane

The Cartesian coordinate plane is a foundational concept in mathematics. It consists of two perpendicular axes—the horizontal axis, known as the x-axis, and the vertical axis, called the y-axis. In the context of the given double integral, we use a similar plane labeled as the t-axis and the u-axis, which together form the tu-plane. Each point on this plane is represented by a pair of coordinates, \(t, u\). The tu-plane helps in visualizing the region over which the integral is evaluated. When solving problems involving double integrals, it's crucial to understand how the region of integration is mapped onto this coordinate plane. The integration bounds provided, such as \(-\frac{\pi}{3}\) to \(\frac{\pi}{3}\) for \t\ and \(0\) to \(\sec t\) for \u\, help to define this region geometrically. This visualization not only simplifies the problem but also gives insights into how functions behave in specific areas of the plane.

Defining the Region of Integration

The region of integration is the area over which we evaluate the double integral. For this specific problem, the given bounds for \(t\) and \(u\) uniquely determine a region on the \tu\-plane. To understand it, think of:

• The limits for \(t\) from \(-\frac{\pi}{3}\) to \(\frac{\pi}{3}\) determine a horizontal slice on the plane.
• For each value of \(t\), the \(u\) value ranges from \(0\) to \(\sec t\), forming vertical boundaries.

These bounds sketch out a region that forms a band above \(u = 0\) and below the curve \(u = \sec t\). To visualize, plot the curve of \(u = \sec t\) within the specified \(t\) interval. This can help in verifying that the defined area accurately represents where integration takes place. Accurate sketching aids greatly in understanding how the function \(3 \cos t\) accumulates its values over this specified region.

Evaluation of Integrals

Evaluating double integrals involves two steps: first, the inner integral and then the outer integral. In our example:

Step 1: Evaluate the Inner Integral

For a given value of \t\, the inner integral \( \int_{0}^{\sec t} 3 \cos t \, du \) is solved. Here, \(3 \cos t\) is constant with respect to \u\, making the integral simple:\[3 \cos t \int_{0}^{\sec t} 1 \, du = 3 \cos t [u]_{0}^{\sec t}\]This means multiplying \(3 \cos t\) by the change in \(u\), \(\sec t - 0\). Substituting \(\sec t = \frac{1}{\cos t}\), simplifies to \(3\).

Step 2: Evaluate the Outer Integral

The outer integral is \(\int_{-\pi / 3}^{\pi / 3} 3 \, dt\). As \(3\) is a constant, it simplifies to:\[3 \int_{-\pi / 3}^{\pi / 3} 1 \, dt = 3 \left[t\right]_{-\pi / 3}^{\pi / 3}\]Calculate the definite integral by evaluating at \(\frac{\pi}{3}\) and \(-\frac{\pi}{3}\), giving \(3 \left( \frac{2\pi}{3} \right) = 2\pi\).Ultimately, mastering how to carry out these integrals step by step will allow you to tackle more complex regions and functions with confidence.