Problem 16 - Thomas Calculus

Question

Set up the iterated integral for evaluating \(\iiint_{D} f(r, \theta, z) d z r d r d \theta\) over the given region \(D .\) \(D\) is the right circular cylinder whose base is the circle \(r=3 \cos \theta\) and whose top lies in the plane \(z=5-x .\)

Step-by-Step Solution

Answer

Set up the iterated integral as \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{3\cos\theta} \int_{0}^{5 - r\cos\theta} f(r, \theta, z) \, dz \, r \, dr \, d\theta\).

Step 1: Identify Region D in Polar Coordinates

The base of the cylinder is defined by the circle equation \(r = 3\cos\theta\) in the plane. This is a polar equation, where \(r\) must be non-negative, thus \(0 \leq r \leq 3\cos\theta\). The angle \(\theta\) ranges over the interval \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\) to cover the circle.

Step 2: Determine the Limits for z

The top of the cylinder is given by the plane \(z = 5 - x\). In polar coordinates, this becomes \(z = 5 - r\cos\theta\). Since the base lies in the xy-plane, the limits for \(z\) are \(0 \leq z \leq 5 - r\cos\theta\).

Step 3: Set Up the Iterated Integral

Using the limits established, the iterated integral over region \(D\) is set up as follows: \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{3\cos\theta} \int_{0}^{5 - r\cos\theta} f(r, \theta, z) \, dz \, r \, dr \, d\theta\). This correctly reflects the region of integration in cylindrical coordinates.

Key Concepts

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system that represent a point in the plane by an angle and a distance from a reference point. This system uses \(r\), the radial distance from the origin, and \(\theta\), the angle measured from the positive x-axis, to describe a point. This approach simplifies the process of working with circles and other shapes where symmetry around a point is involved.
Using polar coordinates, the equation of a circle centered at the origin becomes much simpler, as seen in the exercise where the circle is represented as \(r = 3\cos\theta\). Here, for different angles \(\theta\), the radius \(r\) changes, tracing out a circle.
Conversions between polar and rectangular coordinates are valuable:

• The x-coordinate can be found using \(x = r\cos\theta\)
• The y-coordinate using \(y = r\sin\theta\)

These conversions are particularly handy when moving between calculations in polar and rectangular systems.

Iterated Integrals

Iterated integrals allow us to evaluate multiple integrals in sequence, dealing with one variable at a time. This technique is essential for calculating volumes under surfaces defined by functions of two or more variables.
The iterated integral is performed by integrating first with respect to one variable, then with respect to the next, continuing until all variables are integrated. In the solution, this approach helps in evaluating the multiple integral for the given function over the region \(D\).

When setting up an iterated integral, order matters as different orders of integration can lead to different results. In this exercise, the integration is done in the order \(dz\), \(rdr\), \(d\theta\), reflecting the change in variables from the innermost to the outermost.

Cylindrical Coordinates

Cylindrical coordinates extend polar coordinates by adding a third coordinate, \(z\), to represent height, which makes them extremely useful for three-dimensional problems involving symmetry around a central axis.
In cylindrical coordinates, a point in space is represented by \( (r, \theta, z) \). The conversion from cylindrical to rectangular coordinates mirrors that of polar coordinates with the addition of \(z\):

• \(x = r\cos\theta\)
• \(y = r\sin\theta\)
• \(z = z\)

Using cylindrical coordinates in the problem, we describe a right circular cylinder where the base is defined by \(r = 3\cos\theta\), and the height of the cylinder is determined by \(z = 5 - r\cos\theta\). These expressions define the limits for each variable in the iterated integral.

Limit of Integration

The limit of integration defines the range over which the integration occurs for each variable. It ensures that the integral covers the entire region of interest.
For our exercise, the limits were determined by examining the boundaries of the region \(D\). The inner limits are set by the height of the cylinder, \(0 \leq z \leq 5 - r\cos\theta\), ensuring that integration covers the vertical span correctly.
Next, the limits for \(r\) are provided by the circle equation, \(0 \leq r \leq 3\cos\theta\), which encompasses the radial extent of the region.
The outermost limits, \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), guarantee that the full angular section of the circle is explored. This systematic approach ensures the iterated integral fully encompasses the desired region.