Problem 65 - 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 solid cylinder \(r \leq 2\) that lies between the cones \(\varphi=\pi / 3\) and \(\varphi=2 \pi / 3\)

Step-by-Step Solution

Answer

The volume of the solid is \(V = 4\ln\left(2+\sqrt{3}\right)\).

Step 1: Determine the limits of integration

The given cylinder has a radius \(r=2\). The two cones are defined by the angles \(\varphi=\pi / 3\) and \(\varphi=2 \pi / 3\). In cylindrical coordinates, the volume element is given by \(dV = r\,dr\,d\varphi\,dz\). Here, \(r\) will vary from \(0\) to \(2\), \(\varphi\) will vary from \(\pi / 3\) to \(2\pi / 3\), and \(z\) will vary from \(0\) to the height of the cylinder (which will depend on \(r\) and \(\varphi\)).

Step 2: Calculate the height of the cylinder

From the angle \(\varphi = \pi / 3\), we can find the height of the cylinder by using basic trigonometry. For any point on the cone, \(\tan(\varphi) = z/r\). So, we get \(z = r\tan(\varphi)\). Plugging in the values for \(r\) and \(\varphi\), we have \(z = 2\tan(\pi / 3) = 2\sqrt{3}\).

Step 3: Set up the triple integral

The triple integral for the volume of the solid is given by: $$V = \int\int\int_{R}r\,dr\,d\varphi\,dz$$ We set up the triple integral using the previously found limits of integration and the height of the cylinder: $$V = \int_{\pi / 3}^{2\pi / 3} \int_{0}^{2} \int_{0}^{2\tan(\varphi)} r\,dz\,dr\,d\varphi$$

Step 4: Evaluate the triple integral

First, integrate with respect to \(z\): $$V = \int_{\pi / 3}^{2\pi / 3} \int_{0}^{2} \left[ rz \right]_{0}^{2\tan(\varphi)} dr\,d\varphi$$ $$V = \int_{\pi / 3}^{2\pi / 3} \int_{0}^{2} 2r\tan(\varphi) dr\,d\varphi$$ Next, integrate with respect to \(r\): $$V = \int_{\pi / 3}^{2\pi / 3} \left[ r^2\tan(\varphi) \right]_{0}^{2} d\varphi$$ $$V = \int_{\pi / 3}^{2\pi / 3} 4\tan(\varphi) d\varphi$$ Finally, integrate with respect to \(\varphi\): $$V = \left[ 4\ln|\sec(\varphi) + \tan(\varphi)| \right]_{\pi / 3}^{2 \pi / 3}$$ $$V = 4(\ln|\sec(2\pi / 3) + \tan(2\pi / 3)| - \ln|\sec(\pi / 3) + \tan(\pi / 3)|)$$ $$V = 4\ln\left|\frac{\sec(2\pi / 3) + \tan(2\pi / 3)}{\sec(\pi / 3) + \tan(\pi / 3)}\right|$$ $$V = 4\ln\left(2+\sqrt{3}\right)$$ So, the volume of the solid is \(V = 4\ln\left(2+\sqrt{3}\right)\).

Key Concepts

Triple Integral

A triple integral is an extension of the concept of integration into three dimensions. It is used to calculate volumes and other properties of three-dimensional regions. In the context of the given problem, the triple integral is used to find the volume of a solid bounded by a cylinder and two cones. The integral is set up by specifying functions that represent the region of interest and the differential volume element, which in cylindrical coordinates is expressed as \(dV = r \, dr \, d\varphi \, dz\).
This differential element consists of three parts: the radius \(r\), the angle \(\varphi\), and the height \(z\). When these components are integrated over specified limits, they provide the total volume. The choice of cylindrical coordinates simplifies handling problems with cylindrical symmetry, helping to set up and solve the triple integral efficiently.

Volume Calculation

Volume calculation using integration involves summing up infinitely small volume elements \(dV\) to find the total volume of the object or region. In this exercise, the triple integral is evaluated to calculate the volume of the solid described by the cylinder and two cones.
The integral is structured such that \(r\) varies from 0 to the cylinder's radius, \(\varphi\) spans between the two angles defining the cones, and \(z\) climbs from the base of the cylinder up to a height dependent on \(\varphi\).
The process of integration proceeds through three layers:

• With respect to \(z\), considering its dependency on \(\varphi\)
• With respect to \(r\), covering the radial distance within the cylinder
• Finally, with respect to \(\varphi\), capturing the angular sweep between the cones

Each integration step builds upon the last, ultimately summing to give the entire volume.

Integration Limits

The integration limits specify the bounds within which the variables \(r\), \(\varphi\), and \(z\) vary when solving the triple integral. These limits are crucial as they confine the integration to the region of interest and ensure an accurate volume calculation.
For the cylinder and cone problem, the integration limits are as follows:

• \(r\) ranges from 0 to 2, indicating the radial extent of the solid
• \(\varphi\) spans from \(\pi/3\) to \(2\pi/3\), representing the angles between the two cones
• \(z\) goes from 0 to the surface defined by the cone, given by the equation \(z = r \tan(\varphi)\)

These boundaries ensure that the calculation remains enclosed within the correct geometric space of interest, between the defining cones and cylinder.

Cone and Cylinder Interaction

Understanding the interaction between a cone and a cylinder is essential for correctly setting up our integration problem. The cylinder provides a constant boundary in the radial direction, described by \(r \leq 2\).
The interaction with the cones is dictated by the angles \(\varphi = \pi/3\) and \(\varphi = 2\pi/3\). These cones intersect the cylinder and define surfaces that converge, thus forming a wedge-shaped region within the cylinder.
The angles \(\varphi\) are key to understanding how the cones intersect with the cylinder, as they determine the slope of the cone's surface. The \'interaction\' refers to the spatial region where the surfaces of the cones and cylinder overlap, creating the solid whose volume we're tasked to find. The cylindrical coordinates make it easier to describe this region's shape and to perform the volume integration accurately.