Question

(a) Using spherical coordinates, find the volume cut from the ball \(r \leq a\) by the cone \(\theta=\alpha<\pi / 2\). (b) Show that the \(z\) coordinate of the centroid of the volume in (a) is given by the formula \(\bar{z}=3 a(1+\cos \alpha) / 8\).

Short answer

a) \frac{2\pi a^3}{3} (1 - \cos \alpha)b) \bar{z} = \frac{3 a (1 + \cos \alpha)}{8}

Step-by-step solution

Understand Spherical Coordinates

Conversions and volume element in spherical coordinates: - \(x = r \sin \theta \cos \phi\)- \(y = r \sin \theta \sin \phi\)- \(z = r \cos \theta\)The volume element is \(dV = r^2 \sin \theta \ dr \ d\theta \ d\phi\).

Limits of Integration for Volume in Part (a)

Identify the limits of integration for the given volume:- \(r\) ranges from 0 to \(a\)- \(\theta\) ranges from 0 to \(\alpha\) (since \ \theta \ is limited by \ \alpha \ in the problem)- \ \phi\ ranges from 0 to \2\pi\ (full rotation around the z-axis)

Set Up Integral for Volume

Set up the triple integral for the volume:\[V = \int_{0}^{2\pi} \int_{0}^{\alpha} \int_{0}^{a} r^2 \sin \theta \ dr \ d\theta \ d\phi\]

Evaluate the Integral

Compute the integrals step-by-step:\[V = \int_{0}^{2\pi} \int_{0}^{\alpha} \int_{0}^{a} r^2 \sin \theta \ dr \ d\theta \ d\phi\]1. Integrate with respect to \(r\): \[\int_{0}^{a} r^2 \ dr = \left[ \frac{r^3}{3} \right]_{0}^{a} = \frac{a^3}{3}\]2. Integrate with respect to \(\theta\): \[\int_{0}^{\alpha} \frac{a^3}{3} \sin \theta \ d\theta = \frac{a^3}{3} \left[ -\cos \theta \right]_{0}^{\alpha} = \frac{a^3}{3} (1 - \cos \alpha)\]3. Integrate with respect to \ \phi: \[\int_{0}^{2\pi} \frac{a^3}{3} (1 - \cos \alpha) \ d\phi = \frac{a^3}{3} (1 - \cos \alpha) \cdot 2\pi = \frac{2\pi a^3}{3} (1 - \cos \alpha)\]The volume is then \[V = \frac{2\pi a^3}{3} (1 - \cos \alpha)\]

Set Up Integral for the z-coordinate of the Centroid

The formula for the z-coordinate of the centroid is: \[\bar{z} = \frac{1}{V} \int_{V} z \ dV\]Since \ z = r \cos \theta \, we need to set up a similar integral for the numerator.

Set Up Numerator Integral for the z-coordinate

Set up the triple integral for the z-coordinate:\[\int_{0}^{2\pi} \int_{0}^{\alpha} \int_{0}^{a} r^3 \cos \theta \sin \theta \ dr \ d\theta \ d\phi\]

Evaluate the Integral for the z-coordinate

Compute the integrals step-by-step for the z-coordinate:1. Integrate with respect to \(r\): \[\int_{0}^{a} r^3 \ dr = \left[ \frac{r^4}{4} \right]_{0}^{a} = \frac{a^4}{4}\]2. Integrate with respect to \(\theta\): \[\int_{0}^{\alpha} \frac{a^4}{4} \cos \theta \sin \theta \ d\theta = \frac{a^4}{4} \left[ \frac{-\cos^2 \theta}{2} \right]_{0}^{\alpha} = \frac{a^4}{8} (1 - \cos^2 \alpha)\]3. Integrate with respect to \ \phi: \[\int_{0}^{2\pi} \frac{a^4}{8} (1 - \cos^2 \alpha) \ d\phi = \frac{a^4}{8} (1 - \cos^2 \alpha) \cdot 2\pi = \frac{\pi a^4}{4} (1 - \cos^2 \alpha)\]

Calculate the z-coordinate of the Centroid

Substitute the values into the centroid formula:\[\bar{z} = \frac{1}{V} \int_{V} z \ dV = \frac{\frac{\pi a^4}{4} (1 - \cos^2 \alpha)}{\frac{2\pi a^3}{3} (1 - \cos \alpha)} = \frac{3 a (1 + \cos \alpha)}{8}\]

Key concepts

spherical coordinates

Spherical coordinates are a way to represent points in three-dimensional space. In contrast to Cartesian coordinates \(x, y, z\), spherical coordinates use three parameters: radius (r), polar angle (\(\theta\)), and azimuthal angle (\(\phi\)). These parameters are defined as follows:
- \(r\): The distance from the origin to the point.
- \(\theta\): The angle between the positive z-axis and the line connecting the origin to the point. This angle ranges from 0 to \(\pi\).
- \(\phi\): The angle between the positive x-axis and the projection of the point onto the xy-plane. This angle ranges from 0 to \(2\pi\).

The conversion from spherical to Cartesian coordinates is given by the equations:
\(x = r \sin \theta \cos \phi\)
\(y = r \sin \theta \sin \phi\)
\(z = r \cos \theta\)

The volume element in spherical coordinates is derived from the Jacobian determinant and is represented as:
\(dV = r^2 \sin \theta \, dr \, d\theta \, d\phi\)

This volume element is crucial in setting up integrals for computing the volume and the centroid in spherical coordinates.

volume integration

Volume integration using spherical coordinates often simplifies the computation of volume for objects with symmetry around a point. In our exercise, we need to find the volume cut from a sphere by a cone.

For this specific example:
- \(r\) ranges from 0 to \(a\) (the radius of the sphere)
- \(\theta\) ranges from 0 to \(\alpha\) (the cone's boundary)
- \(\phi\) ranges from 0 to \(2\pi\) (a full rotation around the z-axis)

Setting up the volume integral, we get:
\(V = \int_{0}^{2\pi} \int_{0}^{\alpha} \int_{0}^{a} r^2 \sin \theta \, dr \, d\theta \, d\phi\)

Solving the integral step-by-step:
1. Integrate with respect to \(r\):
\(\int_{0}^{a} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{0}^{a} = \frac{a^3}{3}\)
2. Integrate with respect to \(\theta\):
\(\int_{0}^{\alpha} \frac{a^3}{3} \sin \theta \, d\theta = \frac{a^3}{3} \left[ -\cos \theta \right]_{0}^{\alpha} = \frac{a^3}{3} (1 - \cos \alpha)\)
3. Integrate with respect to \(\phi\):
\(\int_{0}^{2\pi} \frac{a^3}{3} (1 - \cos \alpha) \, d\phi = \frac{a^3}{3} (1 - \cos \alpha) \, \cdot 2\pi = \frac{2\pi a^3}{3} (1 - \cos \alpha)\)

This gives us the volume:
\(V = \frac{2\pi a^3}{3} (1 - \cos \alpha)\)

centroid calculation

The centroid of a volume is the geometric center of that volume. To find the \(z\)-coordinate of the centroid of the volume, we use the formula:
\(\bar{z} = \frac{1}{V} \int_{V} z \, dV\)

Since \(z = r \cos \theta\), we need to set up a similar triple integral for the numerator:
\(\int_{0}^{2\pi} \int_{0}^{\alpha} \int_{0}^{a} r^3 \cos \theta \sin \theta \, dr \, d\theta \, d\phi\)

Solving this integral step-by-step:
1. Integrate with respect to \(r\):
\(\int_{0}^{a} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{a} = \frac{a^4}{4}\)
2. Integrate with respect to \(\theta\):
\(\int_{0}^{\alpha} \frac{a^4}{4} \cos \theta \sin \theta \, d\theta = \frac{a^4}{4} \left[ \frac{-\cos^2 \theta}{2} \right]_{0}^{\alpha} = \frac{a^4}{8} (1 - \cos^2 \alpha)\)
3. Integrate with respect to \(\phi\):
\(\int_{0}^{2\pi} \frac{a^4}{8} (1 - \cos^2 \alpha) \, d\phi = \frac{a^4}{8} (1 - \cos^2 \alpha) \, \cdot 2\pi = \frac{\pi a^4}{4} (1 - \cos^2 \alpha)\)

Finally, substituting this value and the volume from the previous section into the centroid formula:
\(\bar{z} = \frac{1}{V} \int_{V} z \, dV = \frac{\frac{\pi a^4}{4} (1 - \cos^2 \alpha)}{\frac{2\pi a^3}{3} (1 - \cos \alpha)} = \frac{3 a (1 + \cos \alpha)}{8}\)

This matches the given formula for the \(z\)-coordinate of the centroid.

triple integral

A triple integral is used to compute a volume integral in three-dimensional space. It involves integrating a function over a three-dimensional region. In spherical coordinates, the volume element \(dV\) is expressed as \(r^2 \sin \theta \, dr \, d\theta \, d\phi\).

To solve a triple integral:

• First, identify the region of integration and set the limits of integration for \(r\), \(\theta\), and \(\phi\). These limits depend on the geometry of the region.
• Next, set up the integral in the appropriate coordinate system. In our example, spherical coordinates simplify the computation.
• Perform the integrations step-by-step. Start with the innermost integral and proceed outward.

In our volume calculation example, the integral for the volume was:
\(V = \int_{0}^{2\pi} \int_{0}^{\alpha} \int_{0}^{a} r^2 \sin \theta \, dr \, d\theta \, d\phi\)

For the centroid calculation, the integral was:
\(\int_{0}^{2\pi} \int_{0}^{\alpha} \int_{0}^{a} r^3 \cos \theta \sin \theta \, dr \, d\theta \, d\phi\)

Using spherical coordinates and the triple integral method provides a systematic and efficient way to calculate volumes and centroids in three-dimensional space.