Problem 19 - Thomas Calculus

Question

Each of Exercises \(17-20\) 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 Region of Integration

The integral is given in the form \( \int_{a}^{b} \int_{g(t)}^{h(t)} f(t,u) \, du \, dt \). The bounds for \( t \) are \( -\pi/3 \) to \( \pi/3 \) and for \( u \) are from \( 0 \) to \( \sec t \). This describes a region in the \( tu \)-plane where \( t \) is on the \( x \)-axis and \( u \) is on the \( y \)-axis, confined to the line \( u = \sec t \) within the described bounds.

Step 2: Sketch the Region

To sketch the region, plot the lines \( t = -\pi/3 \) and \( t = \pi/3 \), and the curve \( u = \sec t \) between these bounds. For \( t = 0 \), \( \sec t \) equals 1, and as \( t \) approaches \( \pm\pi/3 \), \( \sec t \) approaches 2. Thus, the sketch spans from \( t = -\pi/3 \) to \( t = \pi/3 \) and \( u \) ranges from \( 0 \) to \( \sec t \) within that interval.

Step 3: Setup Inner Integral

The inner integral is \( \int_{0}^{\sec t} 3 \cos t \, du \). Since \( 3 \cos t \) is constant with respect to \( u \), we can directly integrate with respect to \( u \): \( \int 3\cos t \, du = 3\cos t \cdot u \). Applying the bounds, we get \( 3\cos t \cdot (\sec t - 0) = 3 \cos t \cdot \sec t \).

Step 4: Simplify the Inner Integral Result

We now have \( 3 \cos t \cdot \sec t = 3 \cdot 1 = 3 \) because \( \sec t = 1/\cos t \) simplifies this to 3. Thus, the expression to integrate over \( t \) becomes a constant 3.

Step 5: Evaluate the Outer Integral

The outer integral is now \( \int_{-\pi/3}^{\pi/3} 3 \, dt \). Since this is a constant, it simplifies to \( 3 \times (\pi/3 - (-\pi/3)) = 3 \times (2\pi/3) = 2\pi \).

Step 6: Conclusion

The integral evaluates to \( 2\pi \), which is the result of the integration over the defined region in the Cartesian \( tu \)-plane.

Key Concepts

Iterated Integrals

Iterated integrals are essential tools in multivariable calculus for evaluating integrals over regions in a coordinate plane. Essentially, an iterated integral is a nested integral, where you integrate a function first with respect to one variable, and then integrate the result with respect to another variable. This method allows you to systematically approach more complex integrals over defined regions.
In the exercise discussed, we're dealing with the iterated integral in the form \( \int_{a}^{b} \int_{g(t)}^{h(t)} f(t,u) \, du \, dt \). This involves first integrating the function \( f(t,u) \) with respect to \( u \), within the bounds determined by functions of \( t \), followed by integrating the resultant expression with respect to \( t \).
The iterated integral technique is particularly useful when the region of integration is evident or easily described in one coordinate system, like the Cartesian coordinates used here. In application, such integrals allow us to compute volumes under surfaces or aggregate quantities over a specified area.

Region of Integration

Understanding the region of integration is crucial for solving iterated integrals. It defines the area over which the integration occurs. In this problem, the region is within the \( tu \)-plane, which involves horizontal bounds for \( t \) from \( -\pi/3 \) to \( \pi/3 \), and vertical bounds for \( u \) ranging from \( 0 \) to \( \sec t \).
To visualize this region, imagine drawing vertical lines for \( t = -\pi/3 \) and \( t = \pi/3 \). Between these lines, curve the line \( u = \sec t \) which bounds the top of the region. Importantly, \( \sec t \) becomes larger as \( t \) moves towards \( \pm\pi/3 \), stretching the region vertically.
Sketching the region helps to fully appreciate why the iterated integral is arranged the way it is. Once the bounds of integration are clearly defined on a sketch, the iterative process of solving the integrals becomes much more straightforward, allowing each part of the region to be accounted for in the calculation.

Cartesian Coordinates

Cartesian coordinates are a fundamental concept in calculus and geometry, framing every point in a 2D plane by two values: \( x \) and \( y \), or in this case, \( t \) and \( u \). This coordinate system is perfect for breaking down complex functions and regions into manageable pieces through integrals.
In the given problem, the \( tu \)-plane is essentially a Cartesian plane where each point \( (t, u) \) defines a specific position. Representing the bounds of \( t \) and \( u \) in this plane gives you a clear picture of the space over which you're integrating.
Using Cartesian coordinates simplifies the integration setup because it aligns well with how we naturally visualize and determine lengths and changes in shapes. By framing the integral in the Cartesian plane, it becomes easier to apply methods like iterated integrals to solve problems involving areas, and even more advanced concepts, such as volumes under or between surfaces.