Problem 49 - Thomas Calculus

Question

In Exercises \(49-52,\) use a CAS integration utility to evaluate the triple integral of the given function over the specified solid region. \(F(x, y, z)=x^{2} y^{2} z\) over the solid cylinder bounded by \(x^{2}+y^{2}=1\) and the planes \(z=0\) and \(z=1\)

Step-by-Step Solution

Answer

The triple integral evaluates to \( \frac{\pi}{24} \).

Step 1: Understanding the Region and Function

We need to evaluate the triple integral of the function \( F(x, y, z) = x^2 y^2 z \) over a solid cylinder described by the inequalities \( x^2 + y^2 \leq 1 \), \( 0 \leq z \leq 1 \). This solid region is a vertical cylinder of radius 1, height 1, centered on the z-axis.

Step 2: Setting Up the Integral in Cylindrical Coordinates

Convert the given coordinates to cylindrical coordinates. In these coordinates, \( x = r \cos \theta \), \( y = r \sin \theta \), and \( z = z \) remain the same. The solid region transforms into: \( 0 \leq r \leq 1 \), \( 0 \leq \theta \leq 2\pi \), and \( 0 \leq z \leq 1 \). The integral becomes \[ \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{1} (r^2 \cos^2 \theta)(r^2 \sin^2 \theta)z \cdot r \, dz \, dr \, d\theta \].

Step 3: Simplifying the Integrand

Simplify the integrand: \( (r^2 \cos^2 \theta)(r^2 \sin^2 \theta)z \cdot r = r^5 \cos^2 \theta \sin^2 \theta \, z \). Now, the integral is \[ \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{1} r^5 \cos^2 \theta \sin^2 \theta z \, dz \, dr \, d\theta \].

Step 4: Integrating with Respect to z

Since the expression contains \( z \), we integrate it first: \[ \int_{0}^{1} z \, dz = \left[ \frac{z^2}{2} \right]_{0}^{1} = \frac{1}{2} \]. This simplifies the integral to \[ \frac{1}{2} \int_{0}^{2\pi} \int_{0}^{1} r^5 \cos^2 \theta \sin^2 \theta \, dr \, d\theta \].

Step 5: Integrating with Respect to r

Next, integrate with respect to \( r \): \[ \int_{0}^{1} r^5 \, dr = \left[ \frac{r^6}{6} \right]_{0}^{1} = \frac{1}{6} \]. The integral now becomes \[ \frac{1}{12} \int_{0}^{2\pi} \cos^2 \theta \sin^2 \theta \, d\theta \].

Step 6: Evaluating the Angular Integral

Finally, integrate with respect to \( \theta \): \( \cos^2 \theta \sin^2 \theta = \frac{1}{4} (\sin 2\theta)^2 \). Using the identity \( \int (\sin 2\theta)^2 \, d\theta = \int \frac{1 - \cos 4\theta}{2} \, d\theta \), we get \[ \frac{1}{2} \left( \frac{\theta}{2} - \frac{1}{4} \sin 4\theta \right) \]. Evaluating from \( 0 \) to \( 2\pi \), this results in \( \frac{\pi}{2} \). The integral evaluates to \[ \frac{1}{12} \cdot \frac{\pi}{2} = \frac{\pi}{24} \].

Step 7: Conclusion on the Volume of Integration

The triple integral evaluates to \( \frac{\pi}{24} \), which is the result of integrating the given function over the specified cylindrical region.

Key Concepts

Cylindrical Coordinates

Understanding cylindrical coordinates is essential when dealing with problems involving symmetry around an axis, such as cylindrical regions. These coordinates are made up of:

• Radial distance ( ext): Represented by \(r\), this is the distance of the point from the origin projected onto the xy-plane.
• Angular coordinate ( heta ext): Represented by \(\theta\), this is the angle between the positive x-axis and the line from the origin to the projection of the point on the xy-plane.
• Height ( ext{z}): Represented by \(z\), it shows the height of the point above the xy-plane.

Cylindrical coordinates are especially useful for problems with cylindrical geometries because they align with the symmetry of the problem.
When converting from Cartesian coordinates to cylindrical, the formulas are:

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

This transformation helps streamline the computation of integrals over cylinder-like regions.

Solid Cylinder Region

A solid cylinder region is defined by both its circular base and its height. Imagine it as a soda can:

• The circular base is defined by \(x^2 + y^2 = R^2\), where \(R\) is the radius.
• The height is the difference between the top and bottom planes.

For our problem, the region is described as a vertical cylinder where:

• The base lies on the xy-plane with a radius of 1.
• The height goes from \(z = 0\) to \(z = 1\).

When dealing with triple integrals over such regions, understanding the dimensions and boundaries thoroughly is crucial for setting up the integration limits.

Function Integration

Integration is a mathematical method used to find the total accumulation of a quantity.
When we say 'function integration,' we're usually talking about summing up the values a function provides over a certain region.
In this case, we are integrating the function \(F(x, y, z) = x^2 y^2 z\) over the volume of a cylinder. The process involves:

• Breaking down the function into simpler parts with respect to each variable: convert for each cylindrical coordinate in steps.
• Integrating the innermost variable first (here, \(z\)), then the next (\(r\)), and so on.
• Simplifying the function, particularly providing trigonometric identities for \(\cos\) and \(\sin\) for angular integration.

Function integration tells us how the function accumulates as it moves through the volume, informing us about the various values and constraints at play.

Cylinder Volume Calculation

In the context of triple integrals, calculating the volume of a cylinder can be a straightforward task, once we correctly set up the integration expression.
Our cylinder in this problem has:

• A radius \(r = 1\) and height \(z\) running from 0 to 1.
• We are evaluating how the function behaves in this fixed volume, not just calculating an empty volume.

Volume calculation is methodically done by integrating across each dimension:

• First, the integration is done with respect to \(z\), capturing the vertical extent.
• Next, it integrates \(r\) covering the radial spread.
• Lastly, \(\theta\), capturing complete circular coverage.

This layered approach essentially stacks infinitesimally small slices of volume together to acquire the volume
or yielding a result like \(\frac{\pi}{24}\) for function integration, as found in this specific exercise.