Problem 66 - Calculus Early Transcendentals

Question

Miscellaneous volumes Choose the best coordinate system for finding the volume of the following solids. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. That part of the ball \(\rho \leq 2\) that lies between the cones \(\varphi=\pi / 3\) and \(\varphi=2 \pi / 3\)

Step-by-Step Solution

Answer

Question: Determine the volume of a solid bounded by the intersection of a ball with radius 2 and two cones with azimuthal angles of π/3 and 2π/3 in the spherical coordinate system. Answer: The volume of the given solid is \(\frac{16\pi}{9}\).

Step 1: Set up the integration boundaries

We are given the radial boundary as \(\rho \leq 2\), which implies that the range of the radial distance is \(0 \leq \rho \leq 2\). For the cones, we are given boundaries for the azimuthal angle as \(\varphi = \frac{\pi}{3}\) and \(\varphi = \frac{2\pi}{3}\). These boundaries imply that the azimuthal angle, \(\varphi\), ranges from \(\frac{\pi}{3} \leq \varphi \leq \frac{2\pi}{3}\). For the polar angle, since there is no specific contraint for \(\vartheta\), the full range is considered: \(0 \leq \vartheta \leq 2\pi\).

Step 2: Write down the volume element and integration limits

We begin by writing down the volume element in spherical coordinates: \(dV = \rho^2\sin{\vartheta} d\rho d\vartheta d\varphi\). We integrate \(\rho\) from 0 to 2, \(\vartheta\) from 0 to \(2\pi\), and \(\varphi\) from \(\frac{\pi}{3}\) to \(\frac{2\pi}{3}\).

Step 3: Set up the triple integral

The volume of the solid can be found by setting up the following triple integral: \[ V = \int_{\pi/3}^{2\pi/3} \int_{0}^{2\pi} \int_{0}^{2} \rho^{2} \sin{\vartheta} d\rho d\vartheta d\varphi \]

Step 4: Evaluate the integral with respect to \(\rho\)

Let's first solve the integral with respect to \(\rho\): \[ V = \int_{\pi/3}^{2\pi/3} \int_{0}^{2\pi} \left[\frac{1}{3}\rho^3\right]_{0}^{2} \sin{\vartheta} d\vartheta d\varphi = \frac{8}{3}\sin{\vartheta} \int_{\pi/3}^{2\pi/3} \int_{0}^{2\pi} d\vartheta d\varphi \]

Step 5: Evaluate the integral with respect to \(\vartheta\)

Now that the integral with respect to \(\rho\) is evaluated, we move on to solving for \(\vartheta\): \[ V = \frac{8}{3} \int_{\pi/3}^{2\pi/3} \left[-\cos\vartheta\right]_{0}^{2\pi} d\varphi = \frac{8}{3}\int_{\pi/3}^{2\pi/3} 2d\varphi \]

Step 6: Evaluate the integral with respect to \(\varphi\)

Finally, we solve the integral over the range of \(\varphi\): \[ V = \frac{16}{3} \left[ \varphi\right]_{\pi/3}^{2\pi/3} = \frac{16}{3} \left(\frac{2\pi}{3} - \frac{\pi}{3}\right) \]

Step 7: Calculate the volume

The volume of the solid is given by: \[ V = \frac{16}{3}\left(\frac{\pi}{3}\right) = \frac{16\pi}{9} \] Therefore, the volume of the given solid is \(\frac{16\pi}{9}\).

Key Concepts

Spherical Coordinates

Spherical coordinates are incredibly useful for solving problems in three-dimensional space, especially when dealing with round objects like spheres, cones, or cylinders. They offer a way to describe the position of a point in 3D space with just three values: the radial distance \( \rho \), the polar angle \( \vartheta \), and the azimuthal angle \( \varphi \). The radial distance represents how far the point is from the origin, the polar angle indicates the angle with respect to the positive z-axis, and the azimuthal angle gives the angle of the projection of the point in the x-y plane with respect to the positive x-axis. This coordinate system is particularly efficient for integrating over volumes with spherical symmetry.

Volume Element

When calculating the volume of a shape in three dimensions using spherical coordinates, the volume element \(dV\) is essential. It's the equivalent of a small 'cube' in Cartesian coordinates. In spherical coordinates, the volume element is not a simple cube but an infinitesimal section of a shell defined by \(dV = \rho^2\sin{\vartheta} d\rho d\vartheta d\varphi\). This differential volume takes into account the changing shape of the space as one moves further from the origin or changes angles. The \(\rho^2\sin{\vartheta}\) term accounts for the increased 'size' of our volume element as \(\rho\) increases and the spherical 'layer' gets thicker.

Integration Boundaries

When performing a triple integral in spherical coordinates, you must carefully define the integration boundaries for each of the three variables: \(\rho\), \(\vartheta\), and \(\varphi\). These boundaries dictate the limits between which the integration is performed, effectively defining the volume you're interested in calculating. For a sphere, \(\rho\) typically varies from 0 to the radius of the sphere; \(\vartheta\) goes from 0 to \(2\pi\) to cover the full range of 'longitude'; \(\varphi\) is often integrated between 0 and \(\pi\) to span the 'latitude' from pole to pole. However, if the volume is just a segment of a sphere, the boundaries will be more restrictive, corresponding to those segments or layers.

Azimuthal Angle

The azimuthal angle \(\varphi\) is critical when dealing with volumes in spherical coordinates as it tells us about the 'latitude' of a point. It ranges from 0 at the North Pole (positive z-axis), to \(\pi\) at the South Pole (negative z-axis). When you have a volume that's symmetrical along the poles, like a full sphere, you'd integrate \(\varphi\) over its full range. However, for a solid between two cones, you'd adjust the limits for \(\varphi\) to capture just the portion of the volume that lies between these specific angles, as it's the case in our exercise where the specified boundaries are \(\pi/3\) and \(2\pi/3\).

Solid Volume Calculation

To calculate the volume of a solid using spherical coordinates, we set up a triple integral. This triple integral accounts for the varying 'sizes' of volume elements throughout the solid, integrating over the entire volume one piece at a time. It requires proper boundaries for \(\rho\), \(\vartheta\), and \(\varphi\) corresponding to the specifics of the solid's shape and size. Once the integrals are evaluated sequentially for each variable, you end up with the total volume of the solid. In our case, the solid is part of a ball and the limits for each spherical coordinate help us grasp the solid's shape and size for accurate calculation.