Problem 11 - University Calculus: Early Transcendentals

Question

Let \(D\) be the region bounded below by the plane \(z=0,\) above by the sphere \(x^{2}+y^{2}+z^{2}=4,\) and on the sides by the cylinder \(x^{2}+y^{2}=1 .\) Set up the triple integrals in cylindrical coordinates that give the volume of \(D\) using the following orders of integration. a. \(d z d r d \theta\) b. \(d r d z d \theta\) c. \(d \theta d z d r\)

Step-by-Step Solution

Answer

Set up triple integrals with bounds for integration in cylindrical coordinates.

Step 1: Understand the Region Bounded by the Surfaces

The region D is bounded below by the plane z=0, which is the xy-plane. It is bounded above by the sphere given by \(x^2 + y^2 + z^2 = 4\). We also have a cylindrical boundary given by \(x^2 + y^2 = 1\), which implies that \(r = 1\) in cylindrical coordinates.

Step 2: Convert the Bounds to Cylindrical Coordinates

In cylindrical coordinates, we use \(x = r \cos \theta\), \(y = r \sin \theta\), and \(z = z\). The sphere \(x^2 + y^2 + z^2 = 4\) translates to \(r^2 + z^2 = 4\), hence \(z = \sqrt{4 - r^2}\). The cylinder becomes \(r = 1\). The region is bounded by \(0 \leq r \leq 1\) and \(0 \leq \theta < 2\pi\).

Step 3: Set Up the Integral for a. \(dz\ dr\ d\theta\)

For the integral \(dz\ dr\ d\theta\), the bounds of \(z\) vary based on the value of \(r\). For a given \(r\), \(z\) ranges from the plane (\(z = 0\)) to the sphere (\(z = \sqrt{4-r^2}\)). Thus, the integral is:\[V = \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{\sqrt{4-r^2}} 1 \ dz\ dr\ d\theta\]

Step 4: Set Up the Integral for b. \(dr\ dz\ d\theta\)

For the integral \(dr\ dz\ d\theta\), \(r\) is integrated from the cylinder (\(r = 0\)) to \(r = 1\). The bounds for \(z\) remain the same: \(z\) goes from \(0\) to \(\sqrt{4-r^2}\). Therefore, the integral becomes:\[V = \int_{0}^{2\pi} \int_{0}^{\sqrt{4}} \int_{0}^{r_{max}(z)} 1 \ dr\ dz\ d\theta\]Here, \(r_{max}(z) = \sqrt{4 - z^2}\) for each \(z\).

Step 5: Set Up the Integral for c. \(d\theta\ dz\ dr\)

For the integral \(d\theta\ dz\ dr\), \(\theta\) is between 0 and 2\(\pi\). For a fixed \(z\), \(r\) ranges from 0 to \(\sqrt{4 - z^2}\). Hence, the integral is:\[V = \int_{0}^{2\pi} \int_{0}^{\sqrt{4}} \int_{0}^{\sqrt{4-z^2}} 1 \ dr\ dz\ d\theta\]

Key Concepts

Triple Integrals

Triple integrals extend the concept of double integrals to three dimensions, allowing us to compute quantities such as volume over a three-dimensional region. In the context of cylindrical coordinates, triple integrals are useful for evaluating integrals over regions that have circular symmetry. In cylindrical coordinates, every point in space is represented by three parameters: radius (\( r \)), angle (\( \theta \)), and height (\( z \)). This is especially helpful when the region of integration involves cylinders or other circular structures, as it aligns naturally with the geometry of the space. When we set up a triple integral in cylindrical coordinates, we integrate a function over a region bounded by

• the radius \( r \)
• the angle \( \theta \)
• the height \( z \)

This is achieved by expressing the volume element as \( r \, dr \, d\theta \, dz \). Thus, the integral is not only influenced by the height and angle but also by the distance from the center axis, which requires a multiplication by \( r \) to account for the growing area as \( r \) increases.

Volume Computation

Computing the volume of a region using triple integrals involves setting up an integral that evaluates the entirety of the volume through successive integration. In the case of the given exercise, we deal with a region \( D \) bounded above and below by complex surfaces and on the sides by a cylinder.Triple integrals are evaluated iteratively, where we start from the innermost to the outermost integral, integrating along one dimension at a time:

• The innermost integral sweeps across the dimension corresponding to the direction of integration, (e.g., \( z \) for \( dz \)
• The middle integral accounts for the limitations imposed by the shape across the next dimension, such as \( r \) for \( dr \)
• The outermost integral accommodates the rotation or extension around the third dimension, \( \theta \) for \( d\theta \)

Different orders of integration (like \( dz \, dr \, d\theta \), \( dr \, dz \, d\theta \), \( d\theta \, dz \, dr \)) can simplify computations depending on the geometry of the region. It's pivotal to correctly determine the limits of each integration to ensure accurate computation of the volume.

Calculus Integration Techniques

Various integration techniques are used to solve triple integrals, each catering to different complexities of the region being evaluated. Some common techniques include:

• Change of Variables: Converting rectangular coordinates to cylindrical coordinates simplifies the integral setup and calculations when dealing with circular or cylindrical boundaries.
• Iterative Integration: Integrating step-by-step, where each variable is integrated successively, adjusting bounds based on the variable already integrated.
• Recognizing Symmetry: Identifying symmetrical aspects of the integration region can simplify integration, reducing computational effort.

In our example, recognizing that the problem fits well with cylindrical coordinates makes the integral much more manageable. The cylindrical symmetry allows for straightforward bounds for \( \theta \) and \( r \), turning what would be a more complex integral in Cartesian coordinates into a cleaner calculation. By following the appropriate integration techniques, such as correctly switching coordinate systems and evaluating bounds, one can adeptly solve for volumes using cylindrical coordinates.