Question

Use spherical coordinates to find the indicated quantity. Volume of the solid within the sphere \(x^{2}+y^{2}+z^{2}=16\), outside the cone \(z=\sqrt{x^{2}+y^{2}}\), and above the \(x y\) -plane

Short answer

The volume of the solid is \( \frac{64\sqrt{2}\pi}{3} \).

Step-by-step solution

Convert to spherical coordinates

First, express the given surfaces in spherical coordinates. In spherical coordinates, we have: - The sphere: \[ \rho^2 = 16 \quad \Rightarrow \quad \rho = 4 \] - The cone: \[ z = \sqrt{x^2 + y^2} \quad \Rightarrow \quad \cos\varphi = \sin\varphi \Rightarrow \varphi = \frac{\pi}{4} \] Thus, we need to find the volume above the cone, below the sphere \(\rho = 4\), and above the \(xy\)-plane.

Setup the Volume Integral

Set up the volume integral for the region in spherical coordinates. The limits are: - \(0 \leq \theta \leq 2\pi\)- \(\frac{\pi}{4} \leq \varphi \leq \frac{\pi}{2}\)- \(0 \leq \rho \leq 4\)The volume element in spherical coordinates is \(\rho^2 \sin \varphi\,d\rho\,d\varphi\,d\theta\). Thus, the integral becomes:\[ \int_{0}^{2\pi} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{4} \rho^2 \sin \varphi\,d\rho\,d\varphi\,d\theta \]

Evaluate the Integral with respect to \(\rho\)

First, integrate with respect to \(\rho\):\[ \int_{0}^{4} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_{0}^{4} = \frac{4^3}{3} = \frac{64}{3} \] Insert this result back into the integral:

Evaluate the Integral with respect to \(\varphi\)

Evaluate the second integral with respect to \(\varphi\):\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{64}{3} \sin \varphi \, d\varphi = \frac{64}{3} \left[ -\cos \varphi \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \frac{64}{3} \left(-\cos\frac{\pi}{2} + \cos\frac{\pi}{4}\right) \]\[ = \frac{64}{3} \left( 0 + \frac{\sqrt{2}}{2} \right) = \frac{64\sqrt{2}}{6} = \frac{32\sqrt{2}}{3} \]

Evaluate the Integral with respect to \(\theta\)

Finally, integrate with respect to \(\theta\):\[ \int_{0}^{2\pi} \frac{32\sqrt{2}}{3} \, d\theta = \frac{32\sqrt{2}}{3} \left[ \theta \right]_{0}^{2\pi} = \frac{32\sqrt{2}}{3} \times 2\pi = \frac{64\sqrt{2}\pi}{3} \]

Final Volume Calculation

The volume of the solid is \( \frac{64\sqrt{2}\pi}{3} \). This volume represents the region within the sphere, outside the cone, and above the \(xy\)-plane.

Key concepts

Volume Integration

Volume integration is a technique used to find the volume of a region by setting up an integral that represents the three-dimensional space. It involves breaking the volume into infinitesimally small elements and then summing them up to find the total volume. This approach uses integral calculus and is very useful in applications like physics and engineering.

To tackle a volume integration problem, you'll often start by defining the limits of integration for each coordinate direction, which could be cartesian, cylindrical, or spherical depending on the problem. These limits help in confining the region of interest. By integrating over these limits, you can accumulate the volume of the small elements filling that space.

• The integration may involve one or more variables, leading to single, double, or triple integrals.
• Volume integrals are useful for computing mass, charge, or other properties distributed throughout a volume.

In our problem, we use spherical coordinates to compute the volume of a region defined by a sphere and a cone. This involves setting up and evaluating a triple integral over the specified limits.

Spherical Coordinate System

The spherical coordinate system is a three-dimensional coordinate system that extends the idea of polar coordinates to three dimensions. It's particularly useful for dealing with problems involving spheres or spherical regions since it simplifies the equations of such objects, making calculations easier.

In spherical coordinates, a point is represented by three parameters:

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

The conversion from cartesian to spherical coordinates involves:

• The radial distance \( \rho = \sqrt{x^2 + y^2 + z^2} \).
• The polar angle \( \varphi = \arccos(z/\rho) \).
• The azimuthal angle \( \theta = \arctan(y/x) \).

In our problem, the sphere is simply \( \rho = 4 \) whereas the cone maps to the equality condition \( \cos \varphi = \sin \varphi \), simplifying the problem's geometric constraints.

Triple Integral

A triple integral is an extension of single and double integrals, used to compute volumes and other quantities in three-dimensional space. It is expressed as the integral of a function of three variables over a three-dimensional region. For example, in our exercise, the volume element in spherical coordinates is \[ dV = \rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta \]

The limits for these integrals need to be carefully set based on the region you're integrating over. In the context of our problem:

• The angles have ranges: \( 0 \leq \theta \leq 2\pi \) and \( \frac{\pi}{4} \leq \varphi \leq \frac{\pi}{2} \).
• The radial distance has a range \( 0 \leq \rho \leq 4 \).

Evaluating these integrals gives not only the volume but represents a methodology for handling three-dimensional problems in mathematical analysis.

The steps usually involve performing integration sequentially over each variable, often starting with the innermost integral and working outward. This process can simplify complex calculations considerably.

Calculus Problem Solving

Solving calculus problems often requires a combination of mathematical intuition, problem translation, and technique application. When presented with a problem involving multiple constraints or regions, such as volumes bounded by different surfaces, calculus is a powerful tool for description and solution.

Key skills involved in calculus problem solving include:

• Identifying the appropriate coordinate system for simplifying the problem, such as spherical coordinates for symmetrical bodies.
• Setting up integrals with the right limits, ensuring they strictly encompass the desired region.
• Applying fundamental theorems of calculus, such as the integration of functions over specified limits.
• Computing derivatives and antiderivatives as needed throughout the process.

In our example, we explore volume calculation in a region described by a sphere and a cone. The systematic breakdown into steps like converting coordinates, defining integration limits, and evaluating integrals helps students master complex calculus concepts effectively.