Problem 24 - Thomas Calculus

Question

Evaluate the cylindrical coordinate integrals in Exercises \(23-28\) $$ \int_{0}^{2 \pi} \int_{0}^{3} \int_{r^{2} / 3}^{\sqrt{18-r^{2}}} d z r d r d \theta $$

Step-by-Step Solution

Answer

The value of the integral is \(12\pi\).

Step 1: Understand the Integral Limits

This integral is given in cylindrical coordinates. The limits for each variable are provided: \( \theta \) ranges from \(0\) to \(2\pi\), \(r\) from \(0\) to \(3\), and \(z\) from \(\frac{r^2}{3}\) to \(\sqrt{18-r^2}\). First, we need to understand how these limits relate to the volume being integrated over.

Step 2: Set Up the Integral

The integral is set up as \( \int_{0}^{2 \pi} \int_{0}^{3} \int_{r^{2} / 3}^{\sqrt{18-r^{2}}} rz \, dz \, dr \, d\theta \). The integration is done by evaluating from innermost to outermost: first \(z\), then \(r\), and finally \(\theta\).

Step 3: Evaluate the Integral with Respect to z

Integrate with respect to \(z\) within the given limits \(\frac{r^2}{3}\) to \(\sqrt{18-r^2}\): \[ \int_{r^2/3}^{\sqrt{18-r^2}} dz = \left[z\right]_{r^2/3}^{\sqrt{18-r^2}} = \sqrt{18-r^2} - \frac{r^2}{3} \]

Step 4: Integrate the Result with Respect to r

Now substitute the result from Step 3 into the integral with \(r\): \[ \int_{0}^{3} r \left(\sqrt{18-r^2} - \frac{r^2}{3}\right) \, dr \]. This needs to be split into two separate integrals: \( \int_{0}^{3} r \sqrt{18-r^2} \, dr \) and \( -\int_{0}^{3} \frac{r^3}{3} \, dr \).

Step 5: Solve the Integral \(\int r \sqrt{18 - r^2} dr\)

Make a substitution \(u = 18 - r^2\), leading to \(-du = 2r \, dr\) or \(dr = \frac{-du}{2r}\). Adjust the integral: \[ \int r \sqrt{u} \frac{-du}{2r} = -\frac{1}{2} \int \sqrt{u} \, du = -\frac{1}{2} \times \frac{2}{3} u^{3/2} + C \]. Back-substituting, the result becomes \(-\frac{1}{3} (18-r^2)^{3/2}\). Evaluate from \(0\) to \(3\).

Step 6: Solve the Integral \(\int \frac{r^3}{3} dr\)

The integral evaluates to \(\frac{1}{3} \int r^3 \, dr\). Integrating, we get \(\frac{1}{3} \left(\frac{r^4}{4}\right)\) evaluated from \(0\) to \(3\), resulting in \(\frac{1}{4} r^4\) evaluated at points.

Step 7: Evaluate and Combine the Integers in r

The evaluated result for each portion is combined: \[-\frac{1}{3} [(18-3^2)^{3/2} - (18-0^2)^{3/2}] - [\frac{1}{4} (3^4 - 0^4)] = \]. Calculate these values and check simplification.

Step 8: Integrate Over \(\theta\)

Integrate the resulting function with respect to \(\theta\) by multiplying by the range of \(\theta\) (\(2\pi - 0 = 2\pi\)) and apply to the result from Step 7.

Key Concepts

Triple Integral

A triple integral is a way to integrate over a three-dimensional region. In the given problem, the integral is set up in cylindrical coordinates, which simplifies problems involving circular symmetry. A triple integral in cylindrical coordinates has three parts: the integral over the angular coordinate \( \theta \), the radial coordinate \( r \), and the height \( z \).

The structure of a triple integral is crucial because it helps determine the volume or mass in a three-dimensional space, depending on the integrand. In this problem, the integrand is \( rz \), and we are integrating over \( \theta \) from \( 0 \) to \( 2\pi \), \( r \) from \( 0 \) to \( 3 \), and \( z \) from \( \frac{r^2}{3} \) to \( \sqrt{18-r^2} \). This setup points to finding a volume that combines radial and vertical limits, all bound by the surfaces described by \( z \).

In summary, understanding the order and limits of integration is fundamental to successfully evaluating a triple integral, especially in a cylindrical context where conversion from Cartesian coordinates may also be necessary.

Volume Integration

Volume integration involves finding the volume of a three-dimensional region, typically using integrals in either cylindrical or Cartesian coordinates. By setting up a volume integral with the correct limits and integrand, you can evaluate the space within certain boundaries.

Consider the given exercise: each limit defines a part of the volume over which we are integrating. The integration of \( z \), from \( \frac{r^2}{3} \) to \( \sqrt{18-r^2} \), defines the vertical thickness above the radial section. The multiplication by the original integrand \( rz \) accounts for the density of each infinitesimally small volume element within this three-dimensional space.

Volume integration can be particularly useful in physics and engineering when dealing with irregular shapes or objects defined by more than one boundary, as seen with the limits in this exercise. Understanding these concepts helps to approach complex geometries that cannot be solved with simple geometric formulas.

Integration in Polar Coordinates

Integration in polar coordinates often simplifies problems with circular or cylindrical symmetry by transforming them from the common Cartesian system. This approach is widely applicable in various fields because it aligns the coordinate system with the symmetry present in a problem.

The conversion to polar coordinates, and thus cylindrical integration, is accomplished by substituting \( x = r\cos\theta \) and \( y = r\sin\theta \). The integral then includes an additional \( r \) factor to account for the change of area element from \( dx\, dy \) to \( r\, dr d\theta \).

In this specific integration, the factor \( r \) ensures the elements are adjusted for the radial nature of the coordinate system. It's crucial for students to grasp that integration limits in polar coordinates will typically span angles and radii, rather than direct dimensions like length or height. Appreciating these adjustments is key to mastering integration in cylindrical and polar coordinates, paving the way for tackling more advanced calculus problems.