Problem 40 - University Calculus: Early Transcendentals

Question

Let \(D\) be the region in the first octant that is bounded below by the cone \(\phi=\pi / 4\) and above by the sphere \(\rho=3 .\) Express the volume of \(D\) as an iterated triple integral in (a) cylindrical and (b) spherical coordinates. Then (c) find \(V\).

Step-by-Step Solution

Answer

The volume \( V \) is approximately \( 5.560 \).

Step 1: Understanding the Boundaries

We need to express the volume of the region \( D \) between the cone \( \phi = \pi/4 \) and the sphere \( \rho = 3 \) in the first octant. The cone describes a surface in spherical coordinates where \( \phi \) is the angle between the position vector and the vertical axis. The sphere defines a boundary at a constant radius \( \rho = 3 \).

Step 2: Setting up Cylindrical Coordinates

In cylindrical coordinates, we use \( (r, \theta, z) \). Here, we identify that the cone \( \phi = \pi/4 \) translates to \( z = r \). The sphere \( \rho = 3 \) translates to \( x^2 + y^2 + z^2 = 9 \) or \( r^2 + z^2 = 9 \). Our limits will be for \( 0 \leq \theta \leq \pi/2 \), \( 0 \leq r \leq 3 \), and \( r \leq z \leq \sqrt{9 - r^2} \).

Step 3: Cylindrical Coordinates Triple Integral

The volume \( V \) in cylindrical coordinates is expressed as:\[ V = \int_{0}^{\pi/2} \int_{0}^{3} \int_{r}^{\sqrt{9 - r^2}} r \, dz \, dr \, d\theta \]

Step 4: Setting up Spherical Coordinates

In spherical coordinates \( (\rho, \theta, \phi) \), the cone \( \phi = \pi/4 \) is simply a boundary for \( \phi \), while the sphere boundaries are \( \rho = 3 \). The limits for \( D \) are \( 0 \leq \theta \leq \pi/2 \), \( 0 \leq \phi \leq \pi/4 \), and \( 0 \leq \rho \leq 3 \).

Step 5: Spherical Coordinates Triple Integral

The volume \( V \) in spherical coordinates can be calculated as:\[ V = \int_{0}^{\pi/2} \int_{0}^{\pi/4} \int_{0}^{3} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]

Step 6: Evaluating Spherical Integral Step 1

First, compute the innermost integral with respect to \( \rho \):\[ \int_{0}^{3} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^3 = 9 \]

Step 7: Evaluating Spherical Integral Step 2

Then, compute the integral with respect to \( \phi \):\[ \int_{0}^{\pi/4} 9 \sin \phi \, d\phi = -9 \cos \phi \Big|_0^{\pi/4} = -9 \left( \frac{1}{\sqrt{2}} - 1 \right) \]

Step 8: Evaluating Spherical Integral Step 3

Finally, compute the integral with respect to \( \theta \):\[ \int_{0}^{\pi/2} -9 \left( \frac{1}{\sqrt{2}} - 1 \right) \, d\theta = -9 \left( \frac{1}{\sqrt{2}} - 1 \right) \cdot \frac{\pi}{2} \]

Step 9: Compute Final Volume

Evaluating the constants and simplifying gives the volume \( V \) of region \( D \). The final expression evaluates to:\[ V = 9 \left(1 - \frac{1}{\sqrt{2}}\right) \frac{\pi}{2} \approx 5.560 \]

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates provide a way to describe points in three-dimensional space using a combination of polar coordinates in the xy-plane and a height (z) coordinate. In this system, the position of a point is specified by:

• \( r \) - the radial distance from the origin to the projection of the point on the xy-plane.
• \( \theta \) - the angle between the positive x-axis and the line connecting the origin to the projection of the point on the xy-plane.
• \( z \) - the height above the xy-plane.

Unlike Cartesian coordinates, cylindrical coordinates are particularly useful for problems with symmetry around a central axis. To solve our problem using cylindrical coordinates, we first translate the given boundaries for the cone and the sphere into equations involving \( r, \theta, \) and \( z \). The cone described by \( \phi = \pi/4 \) transforms into the equation \( z = r \). Next, the sphere defined by \( \rho = 3 \) translates to \( x^2 + y^2 + z^2 = 9 \), or equivalently \( r^2 + z^2 = 9 \) in cylindrical terms. The integration bounds for \( \theta \) and \( r \) are chosen considering we are working in the first octant, which means \( 0 \leq \theta \leq \frac{\pi}{2} \) and \( 0 \leq r \leq 3 \). The limits for \( z \) are between the cone surface \( z = r \) and the sphere surface \( z = \sqrt{9 - r^2} \). These boundaries convert into our triple integral expression to calculate the volume.

Spherical Coordinates

Spherical coordinates involve a different approach where each point in space is identified by three parameters:

• \( \rho \) - the radial distance from the origin to the point.
• \( \theta \) - the azimuthal angle in the xy-plane from the positive x-axis.
• \( \phi \) - the polar angle from the positive z-axis.

This system is especially helpful for dealing with symmetric problems involving spheres and cones, as it simplifies the equations governing such geometries. In our exercise, we apply spherical coordinates to determine the volume between a cone and a sphere. The cone \( \phi = \pi/4 \) sets a boundary for the angle \( \phi \). The sphere, indicated by \( \rho = 3 \), provides a radial boundary for our integration. Thus, our limits are \( 0 \leq \theta \leq \frac{\pi}{2} \), \( 0 \leq \phi \leq \pi/4 \), and \( 0 \leq \rho \leq 3 \). When setting up the triple integral for spherical coordinates to find the volume \( V \), we integrate \( \rho^2 \sin \phi \) within these limits. Each segment, from \( \rho \) to \( \theta \), contributes to the total volume by incorporating the spherical differential element \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \).

Volume Calculation

Calculating the volume, whether in cylindrical or spherical coordinates, involves carefully setting up and evaluating a triple integral. Each method simplifies integration through the axes or angles linked with the problem geometry. In cylindrical coordinates, we integrate the volume element \( r \, dz \, dr \, d\theta \) over the specified limits. The specific bounds for each variable ensure that the entire region between the cone and the sphere in the first octant is covered.
After setting up the integral, it involves calculating:

• The innermost integral with respect to \( z \), representing the height between the cone and the sphere.
• Next, the integral with respect to the radial distance \( r \), captures the change in size as we move away from the z-axis.
• Finally, integrating over \( \theta \) accounts for the region's full sweep around the z-axis within the first octant.

In the spherical coordinate system, we calculate volume using \( \rho^2 \sin \phi \), integrating first with respect to \( \rho \), then \( \phi \), and finally \( \theta \). The volume obtained in this manner roughly evaluates to 5.560 cubic units. This final outcome allows us to appreciate how each coordinate system, depending on symmetry and boundary conditions, provides a powerful tool for tackling complex three-dimensional volume problems.