Problem 31 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Describe the Region of Integration

The given integral is \( \int_{-\pi/3}^{\pi/3} \int_{0}^{\sec t} 3 \cos t \, du \, dt \). This represents a region in the \( tu \)-plane where \( t \) ranges from \(-\pi/3\) to \( \pi/3 \) and for each \( t \), \( u \) ranges from 0 to \( \sec t \). The upper bound \( \sec t \) indicates that at each \( t \), \( u \) reaches up to the secant line. The region is bounded by these curves on the plane.

Step 2: Sketch the Region in the tu-Plane

In the \( tu \)-plane, plot the lines \( t = -\pi/3 \) and \( t = \pi/3 \). These vertical lines are the boundaries for \( t \). For \( u \leq \sec t \), plot \( u = \sec t \) as a curve where the secant function goes from \( 1 \) at \( t=0 \) to higher values towards the boundaries of \( t = -\pi/3 \) and \( t = \pi/3 \). The region is thus the area between these curves from \( u = 0 \) to \( u = \sec t \) for the given range of \( t \).

Step 3: Evaluate the Inner Integral with respect to u

Evaluate \( \int_{0}^{\sec t} 3 \cos t \, du \). Since \( 3 \cos t \) is a constant with respect to \( u \), the result of the inner integral is \( 3 \cos t (u) \) evaluated from 0 to \( \sec t \), which simplifies to \( 3 \cos t \sec t \).

Step 4: Simplify the Expression

Simplify the expression obtained in Step 3. Use the identity \( \sec t = \frac{1}{\cos t} \) to rewrite the integral result as \( 3 \). Thus, the integral simplifies to the constant value 3 across the interval of \( t \).

Step 5: Evaluate the Outer Integral with respect to t

Integrate \( 3 \) with respect to \( t \) over \(-\pi/3\) to \( \pi/3 \). The integral becomes: \[ \int_{-\pi/3}^{\pi/3} 3 \, dt = 3 \left[ t \right]_{-\pi/3}^{\pi/3} = 3 \left( \pi/3 - (-\pi/3) \right) = 3 \cdot \left( 2\pi/3 \right) \].

Step 6: Final Evaluation

Multiply the constant 3 by the evaluated integral \( (2\pi/3) \), resulting in \( 2\pi \). This is the value of the doubled integral over the specified region of the \( tu \)-plane.

Key Concepts

Region of Integration

In double integration, the region of integration specifies the area over which we are evaluating the integral. It is crucial to clearly understand this region as it defines the limits for the integrals. In our exercise, the region of integration is defined in the

• horizontal direction by \( t \) ranging from \(-\pi/3\) to \(\pi/3\)
• vertical direction by \( u \) ranging from 0 to \(\sec t\)

This region can be visualized on the Cartesian coordinate plane, where each pair \((t, u)\) falls within these limits. Understanding this area is key to setting up the double integral correctly.

Cartesian Coordinate Plane

The Cartesian coordinate plane is a fundamental concept in mathematics that allows us to graphically represent equations and regions. It comprises two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). In the context of this exercise,

• we use \((t, u)\) as our variables in a plane similar to the \((x, y)\) that we might typically encounter

Each point on this plane corresponds to a specific set of values of \(t\) and \(u\), and together, they create a graphical image of the region of integration previously described. This setup is important for visually understanding bounds and regions involved in double integration.

Secant Function

The secant function, denoted as \(\sec t\), is a trigonometric function defined as the reciprocal of the cosine function: \(\sec t = \frac{1}{\cos t}\). This function plays an essential role in determining the upper bound for the inner integral in our problem.

• At \(t = 0\), \(\sec t\) equals 1
• As \(t\) moves towards \(\pm \pi/3\), \(\sec t\) increases exceeding the value of 1

This increasing nature of \(\sec t\) as \(t\) increases or decreases from 0 ensures that the upper boundary of our integral region is dynamic and changes with \(t\). Understanding this is vital for correctly computing the integral.

Inner Integral

The inner integral is the part of a double integral evaluated first, and it is typically with respect to the variable inside the brackets, here denoted as \(u\).
For the given exercise,

• the inner integral is \(\int_{0}^{\sec t} 3 \cos t \, du\)

Since \(3 \cos t\) is a constant for the integration with respect to \(u\),

• the integral simplifies to \(3 \cos t \cdot u\)

evaluated from 0 to \(\sec t\), which simplifies further to \(3\), as \(\sec t \cdot \cos t = 1\). This process is crucial, as solving the inner integral forms part of the path to finding the solution for the outer integral.

Outer Integral

The outer integral is the final step in solving a double integral problem, evaluated after computing the inner integral. In our problem, once the inner integral is simplified to a constant value, the outer integral becomes

• \(\int_{-\pi/3}^{\pi/3} 3 \, dt\)

This simple integration involves multiplying the constant by the length of the interval of integration.
The process calculates the area under a constant function over the specified range and results in

• \(3 \times \left(2\pi/3\right)\)

which simplifies to \(2\pi\). Evaluating the outer integral completes the task of finding the total area or volume under the function across the defined region.