Problem 17 - Calculus Early Transcendentals

Question

Find the volume of the following solids using triple integrals. The solid bounded below by the cone \(z=\sqrt{x^{2}+y^{2}}\) and bounded above by the sphere \(x^{2}+y^{2}+z^{2}=8\)

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(\frac{32\pi^2}{3}\).

Step 1: Change to Cylindrical Coordinates

In cylindrical coordinates, we represent the points in space with the coordinates \((r, \theta, z)\). The conversion between Cartesian and cylindrical coordinates is as follows: \[x = r\cos{\theta}\] \[y = r\sin{\theta}\] \[z = z\]For the given surfaces, we have:\[z = \sqrt{x^2 + y^2} \implies z = r\] \[x^2 + y^2 + z^2 = 8 \implies r^2 \cos^2{\theta} + r^2 \sin^2{\theta} + z^2 = 8\]

Step 2: Set up the triple integral

The next step is to set up the triple integral. For this problem, we will integrate with respect to \(r\), \(\theta\), and \(z\). The bounds for \(\theta\) and \(r\) can be determined by examining the region below the cone and above the xy-plane since these bounds will be the same for the entire solid. The bounds for \(z\) will be determined by the cone and the sphere, which are the lower and upper bounds, respectively. The bounds for \(\theta\) are \(0\) to \(2\pi\). The bounds for \(r\) are \(0\) to \(\sqrt{8}\) since the sphere intersects the xy-plane at \((\sqrt{8}, 0, 0)\). The bounds for \(z\) are \(r\) to \(\sqrt{8-r^2}\), as they are determined by the cone and sphere. The triple integral in cylindrical coordinates becomes: \[\int_{0}^{2\pi}\int_{0}^{\sqrt{8}}\int_{r}^{\sqrt{8-r^2}} rdzdrd\theta\]

Step 3: Evaluate the integral

We will solve the integral step by step. First, we will integrate with respect to \(z\). \[\int_{r}^{\sqrt{8-r^2}} rdz = \Big[rz\Big]_{z=r}^{\sqrt{8-r^2}} = r(\sqrt{8-r^2} - r)\]Next, we integrate with respect to \(r\). \[\int_{0}^{\sqrt{8}} r(\sqrt{8-r^2} - r)dr = \frac{8\pi}{3}\]Finally, we integrate with respect to \(\theta\). \[\int_{0}^{2\pi} \frac{8\pi}{3}d\theta = \frac{32\pi^2}{3}\]The volume of the solid bounded below by the cone and above by the sphere is \(\frac{32\pi^2}{3}\).

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates provide an alternative way to describe points in three-dimensional space, especially when dealing with figures with circular symmetry. This system uses three parameters: radial distance \(r\), angle \(\theta\), and height \(z\). These coordinates are transformations of the familiar Cartesian coordinates \((x, y, z)\). The equations used for transforming are:

• \(x = r \cos{\theta}\)
• \(y = r \sin{\theta}\)
• \(z = z\)

In the context of the given exercise, we used cylindrical coordinates to simplify the solid defined by a cone and a sphere. Here, the equation of the cone \(z = \sqrt{x^2 + y^2}\) converted directly to \(z = r\) simplifying the integration process. Meanwhile, the sphere equation \(x^2 + y^2 + z^2 = 8\) remained largely unchanged except for expressing \(x^2 + y^2\) as \(r^2\). This transformation is beneficial for integrating over circularly symmetric regions.

Volume Calculation

Calculating the volume of a solid using triple integrals involves integrating over a three-dimensional region. The technique is particularly powerful when combined with cylindrical coordinates, as it simplifies dealing with surfaces that exhibit symmetry around an axis. In the given exercise, we set up a triple integral to find the volume of the solid bounded by a cone and a sphere.
The order of integration for this case is \(z\), \(r\), and then \(\theta\). Each integral has its bounds:

• For \(\theta\), the interval is \(0\) to \(2\pi\), because it encompasses the entire circular base of the solid.
• For \(r\), the range is from \(0\) to \(\sqrt{8}\). This accounts for the radius up to where the sphere intersects the plane \(z=0\).
• The \(z\) bounds are from \(r\) to \(\sqrt{8-r^2}\), as defined by the positions of the cone and sphere.

By properly setting these bounds, we integrated step-wise first with \(z\), then \(r\), and finally \(\theta\), ultimately finding that the volume enclosed is \(\frac{32\pi^2}{3}\). This result showcases the effectiveness of triple integrals for solving complex spatial problems.

Boundary Surfaces

Boundary surfaces are essential in understanding the region over which we integrate using cylindrical coordinates. These surfaces define where our integration starts and stops. In the given problem, the boundaries are formed by a cone and a sphere.

• The lower boundary is the cone \(z = \sqrt{x^2 + y^2}\), represented in cylindrical coordinates as \(z = r\). This defines the bottom surface of our solid.
• The upper boundary is the sphere \(x^2 + y^2 + z^2 = 8\), which in cylindrical terms became \(r^2 + z^2 = 8\). Solving for \(z\), the upper limit \(z = \sqrt{8-r^2}\) was determined.

To find the volume of the defined solid, the space between these boundaries is filled by the triple integral. Each boundary plays a critical role in dictating the limits of integration for \(z\), ensuring we integrate over exactly the region bounded by these two surfaces. Through understanding and setting these boundaries, the integral accurately reflects the volume of the solid between the cone and the sphere.