Problem 49 - University Calculus: Early Transcendentals

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 integral evaluates to \( \frac{\pi}{48} \).

Step 1: Understand the Region of Integration

The solid region is a cylinder described by the equation \( x^2 + y^2 = 1 \) with its height extending from \( z = 0 \) to \( z = 1 \). The limits for \( z \) are from 0 to 1, and for the cylindrical base, \( x^2 + y^2 \leq 1 \).

Step 2: Set Up the Integral

The triple integral of \( F(x, y, z) = x^2 y^2 z \) over the cylinder is set up as follows: \(\int_{z=0}^{1} \int_{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{x=-1}^{1} x^2 y^2 z \; dx \, dy \, dz\). This accounts for the boundaries of the solid cylinder. But since the base is circular, we can use polar coordinates for \( x \) and \( y \).

Step 3: Convert to Cylindrical Coordinates

We substitute \( x = r \cos(\theta) \), \( y = r \sin(\theta) \), and \( dx \, dy \) by \( rdrd\theta \) due to the Jacobian determinant. The integral becomes \(\int_{z=0}^{1} \int_{\theta=0}^{2\pi} \int_{r=0}^{1} (r^2 \cos^2(\theta) \cdot r^2 \sin^2(\theta) \cdot z) \cdot r \, dr \, d\theta \, dz\).

Step 4: Simplify the Integral Expression

This integral simplifies to \( \int_{z=0}^{1} \int_{\theta=0}^{2\pi} \int_{r=0}^{1} r^5 z \cos^2(\theta) \sin^2(\theta) \, dr \, d\theta \, dz\). The expression can be further simplified by evaluating each part sequentially starting with \( r \).

Step 5: Integrate with Respect to \( r \)

The integral of \( r^5 \) over \( r \, \) is \(\int_{r=0}^{1} r^5 \, dr = \frac{r^6}{6} \bigg|_{0}^{1} = \frac{1}{6}.\).

Step 6: Integrate with Respect to \( z \)

Now integrate with respect to \( z \): \(\int_{z=0}^{1} z \, dz = \frac{z^2}{2} \bigg|_{0}^{1} = \frac{1}{2}.\)

Step 7: Integrate with Respect to \( \theta \)

Finally, integrate with respect to \( \theta \): \(\int_{\theta=0}^{2\pi} \cos^2(\theta) \sin^2(\theta) \, d\theta.\) Use the identity \( \cos^2(\theta) \sin^2(\theta) = \frac{1}{4} \sin^2(2\theta) \), giving \(\frac{1}{4} \int_{0}^{2\pi} \sin^2(2\theta) \, d\theta = \frac{1}{8} \left[\theta - \frac{1}{2}\sin(4\theta)\right]_{0}^{2\pi} = \frac{\pi}{4}.\).

Step 8: Compute the Final Result

Multiplying the results of all parts together: \(\frac{1}{6} \times \frac{1}{2} \times \frac{\pi}{4} = \frac{\pi}{48}.\) Thus the value of the triple integral is \( \frac{\pi}{48}.\)

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates provide a practical way to describe three-dimensional space, particularly for problems involving cylinders or circular symmetry. This system is an extension of polar coordinates in two-dimensional space by adding a third dimension, the height or axial coordinate.

• Instead of using Cartesian coordinates \(x, y, z\), we express space using the coordinates \(r, \theta, z\).
• Here, \(r\) represents the radial distance from the origin to the projection of the point in the \(xy\)-plane.
• The angle \(\theta\) is measured counterclockwise from the positive \((x)\)-axis.
• The coordinate \(z\) represents the height, just as in Cartesian coordinates.

Using cylindrical coordinates makes it easier to set up integrals over regions with circular bases, such as the cylinder in our original exercise.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by a distance and an angle. As the fundamental base for cylindrical coordinates, understanding polar coordinates is essential.

• In polar coordinates, a point \( (x, y) \) in Cartesian space can be expressed as \( (r, \theta) \).
• The variable \( r \) is the distance from the origin to the point, while \( \theta \) is the angle between the positive \( x \)-axis and the line connecting the origin to the point.

This transformation can be denoted mathematically as \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). These equations are pivotal in converting Cartesian integrals to polar or cylindrical coordinates.

Jacobian Determinant

When transforming coordinates, as from Cartesian to cylindrical, we must ensure that the volume element is accurately represented. This is where the Jacobian determinant becomes crucial.

• The Jacobian determinant helps adjust the volume element from one coordinate system to another.
• For the conversion from Cartesian to cylindrical coordinates, the Jacobian is \( r \).
• It reflects the area scaling factor used in integration.

Thus, when integrating in cylindrical coordinates, the differential area \(dx \, dy\) is transformed into \(r \, dr \, d\theta\). This adjustment ensures that we capture the true size of each infinitesimal area in our integration results.

Cylindrical Integration

Cylindrical integration is straightforward once the transformation from rectangular to cylindrical coordinates is complete. The power of cylindrical integration lies in its ability to simplify problems with symmetrical, circular boundaries.

• Replace Cartesian coordinates \(x, y\) with cylindrical expressions \(r \cos(\theta), r \sin(\theta)\).
• Convert the differential area \(dx \, dy\) to \(r \, dr \, d\theta\) using the Jacobian.
• Integrals are then evaluated over the simpler radial and angular limits.
• This transformation often results in less complex integrals, which are easier to evaluate.

For our exercise, this process took the complex triple integral and transformed it into a series of manageable single-variable integrals.

Multivariable Calculus

Multivariable calculus generalizes the concepts of single-variable calculus to functions of two or more variables. It is the mathematical study underlying our triple integral problem.

• This branch of calculus explores limits, continuity, differentiation, and integration in multi-dimensional settings.
• It offers tools like partial derivatives, multiple integrals, and coordinate transformations.
• For functions of more than one variable, we often need to consider changes in one variable while keeping others constant.

In our exercise, using multivariable calculus allowed us to evaluate the triple integral over the defined solid region, calculating the accumulated continuous sum of the function values throughout the specified volume.