Question

As needed, use a computer to plot graphs and to check values of integrals. Find the volume inside the cone \(z^{2}=x^{2}+y^{2},\) above the \((x, y)\) plane, and between the spheres \(x^{2}+y^{2}+z^{2}=1\) and \(x^{2}+y^{2}+z^{2}=4 .\) Hint: Use spherical coordinates.

Short answer

The volume inside the cone and between the spheres is \(\frac{14\pi}{3} (1 - \frac{1}{\sqrt{2}})\).

Step-by-step solution

- Convert to spherical coordinates

Use spherical coordinates where \(x = \rho \, \text{sin} \theta \, \text{cos} \phi \), \(\, y = \rho \, \text{sin} \theta \, \text{sin} \phi \, \), and \(\, z = \rho \, \text{cos} \theta \). The volume element in spherical coordinates is \(dV = \rho^2 \, \text{sin} \theta \, d\rho \, d\theta \, d\phi\).

- Determine the bounds for \(\rho\)

The volume is bounded by the spheres \(x^2 + y^2 + z^2 = 1\) and \(x^2 + y^2 + z^2 = 4\). Thus, \(\rho\) ranges from 1 to 2.

- Determine the bounds for \(\theta\)

The cone \(z^2 = x^2 + y^2\) translates to \(\text{cos}^2 \theta = \text{sin}^2 \theta \), meaning \(\theta\) ranges from 0 to \(\pi/4\).

- Determine the bounds for \(\phi\)

\(\phi\) represents the azimuthal angle, which ranges from 0 to \(2\pi\).

- Setup the integral

The integral to find the volume is \[V = \int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{1}^{2} \rho^2 \, \text{sin} \theta \, d\rho \, d\theta \, d\phi \]

- Integrate with respect to \(\rho\)

Evaluate the inner integral: \[\int_{1}^{2} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_{1}^{2} = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} \]

- Integrate with respect to \( \theta \)

Evaluate the next integral: \[ \int_{0}^{\pi/4} \frac{7}{3} \, \text{sin} \theta \, d\theta = \frac{7}{3} \left[- \text{cos} \theta \right]_{0}^{\pi/4} = \frac{7}{3} (1 - \frac{1}{\sqrt{2}}) \]

- Integrate with respect to \(\phi\)

Evaluate the outer integral: \[ \int_{0}^{2\pi} \frac{7}{3} (1 - \frac{1}{\sqrt{2}}) \, d\phi = \frac{7}{3} (1 - \frac{1}{\sqrt{2}}) \cdot 2\pi = \frac{14\pi}{3} (1 - \frac{1}{\sqrt{2}}) \]

Key concepts

spherical coordinates

Spherical coordinates are a three-dimensional coordinate system. They are used to define a point in space with three values: radial distance (\(\rho\)), polar angle (\(\theta\)), and azimuthal angle (\(\phi\)). These are helpful for problems involving symmetry such as spheres and cones. The relationship between rectangular coordinates \((x, y, z)\) and spherical coordinates is:

• x = \(\rho \, \text{sin} \theta \, \text{cos} \phi \)
• y = \(\rho \, \text{sin} \theta \, \text{sin} \phi \)
• z = \(\rho \, \text{cos} \theta \)

Understanding these relationships allows you to transform a complex integral into a simpler form. For volume calculations in spherical coordinates, the volume element is \(dV = \rho^2 \, \text{sin} \theta \, d\rho \, d\theta \, d\phi \). This change of variables often simplifies the evaluation.

volume integral

A volume integral helps to find the volume of a three-dimensional region. When using spherical coordinates, the volume integral is written as:

• \[V = \int_{volume} dV = \int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{1}^{2} \rho^2 \, \text{sin} \theta \, d\rho \, d\theta \, d\phi \]

This integral sums up tiny volume elements across the given bounds. The bounds will typically derive from the geometric constraints of the problem, such as the sphere and cone in our example. Each integral is computed step-by-step, respecting the inner integral's bounds while moving outward.

multivariable calculus

Multivariable calculus extends the concepts of single-variable calculus to higher dimensions. It involves several variables and functions that depend on more than one variable. This field covers differentiation and integration in higher dimensions.
In our context, we use triple integrals to compute volume. It involves stepping through each variable in sequence, starting with the innermost until the complete integral is computed. This approach allows tackling complex regions and functions efficiently.

bounds determination

Determining the bounds for the variables is a crucial step in setting up a volume integral. Each variable will have its range dictated by the problem constraints.
In our exercise:

• The radial distance \(\rho\) ranges from 1 to 2, given by the spheres \(x^2 + y^2 + z^2 = 1\) and \(x^2 + y^2 + z^2 = 4\).
• The polar angle \(\theta\) ranges from 0 to \(\pi/4\), derived from the cone's equation \(z^2 = x^2 + y^2\).
• The azimuthal angle \(\phi\) covers a full rotation around the z-axis, thus ranging from 0 to \(2\pi\).

Accurately determining these bounds is essential for setting up and solving the integral correctly.

integral evaluation

Evaluating the integral involves computing each part step-by-step while respecting the determined bounds. Here's how we do it in our example:

• First, integrate with respect to \(\rho\): \[\int_{1}^{2} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_{1}^{2} = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}\]
• Next, integrate with respect to \(\theta\): \[\int_{0}^{\pi/4} \frac{7}{3} \, \text{sin} \theta \, d\theta = \frac{7}{3} \left[- \text{cos} \theta \right]_{0}^{\pi/4} = \frac{7}{3} (1 - \frac{1}{\sqrt{2}})\]
• Finally, integrate with respect to \(\phi\): \[\int_{0}^{2\pi} \frac{7}{3} (1 - \frac{1}{\sqrt{2}}) \, d\phi = \frac{7}{3} (1 - \frac{1}{\sqrt{2}}) \cdot 2\pi = \frac{14\pi}{3} (1 - \frac{1}{\sqrt{2}})\]

Each integration step simplifies the expression further until we get the final volume solution.