Question

Use spherical coordinates to find the volume of the following regions. A ball of radius \(a>0\)

Short answer

Title: Volume of a Ball Using Spherical Coordinates Converting into Spherical Coordinates: To find the volume of a ball with radius a, we first convert the equation of the ball into spherical coordinates (r, θ, φ). The general relationship between rectangular coordinates (x, y, z) and spherical coordinates is given by the equations: - \(x = r \sin{\phi} \cos{\theta}\) - \(y = r \sin{\phi} \sin{\theta}\) - \(z = r \cos{\phi}\) The equation of the ball in spherical coordinates is \(r = a\) Setting up the Triple Integral: We set up the triple integral to calculate the volume using the volume element in spherical coordinates: \(dV = r^2 \sin{\phi} dr d\phi d\theta\). Thus, the triple integral for the volume is: \(V = \int \int \int r^2 \sin{\phi} dr d\phi d\theta\) Determining the Integration Limits: The integration limits are determined by considering the full range of the radius, polar angle, and azimuthal angle to cover the entire volume of the ball. - For radius, we go from \(0\) to \(a\): \(0 \leq r \leq a\) - For the polar angle, we go from \(0\) to \(\pi\): \(0 \leq \phi \leq \pi\) - For the azimuthal angle, we go from \(0\) to \(2\pi\): \(0 \leq \theta \leq 2\pi\) Computing the Triple Integral: We compute the triple integral with the determined limits: \(V = \int_{0}^{a} \int_{0}^{\pi} \int_{0}^{2\pi} r^2 \sin{\phi} dr d\phi d\theta\) We proceed in three steps: integrate with respect to r, then with respect to φ, and finally with respect to θ: 1. \(\int_{0}^{a} r^2 dr =\dfrac{1}{3}r^3 \bigg|_{0}^{a} = \dfrac{1}{3}a^3\) 2. \(\int_{0}^{\pi} \sin{\phi} d\phi = -\cos{\phi}\bigg|_{0}^{\pi} = 2\) 3. \(\int_{0}^{2\pi} d\theta = \theta \bigg|_{0}^{2\pi} = 2\pi\) Finding the Volume: Finally, we multiply the results from the three steps: \(V = \dfrac{1}{3}a^3 \cdot 2 \cdot 2\pi = \dfrac{4}{3}\pi a^3\) Therefore, the volume of a ball with radius a in spherical coordinates is \(\dfrac{4}{3}\pi a^3\).

Step-by-step solution

Converting into Spherical Coordinates

Instead of using rectangular coordinates (x, y, z), we will convert the ball equation into spherical coordinates denoted by (r, θ, φ). In general, the spherical coordinates are related to the rectangular coordinates by the following equations: - \(x = r \sin{\phi} \cos{\theta}\) - \(y = r \sin{\phi} \sin{\theta}\) - \(z = r \cos{\phi}\) Since we are asked to find the volume of a ball of radius a, we know that the equation in spherical coordinates will be in the form: \(r = a\)

Setting up the Triple Integral

Now that we have the equation in spherical coordinates, it's time to set up the triple integral which will give us the volume of the ball. The volume element in spherical coordinates is given by the following equation: \(dV = r^2 \sin{\phi} dr d\phi d\theta\) Setting up the volume integral for the ball, we have: \(V = \int \int \int r^2 \sin{\phi} dr d\phi d\theta\)

Determining the Integration Limits

To find the integration limits, we need to consider the full range of the radius, polar angle, and azimuthal angle to cover the entire volume of the ball. - For radius, we go from \(0\) to \(a\): \(0 \leq r \leq a\) - For the polar angle, we go from \(0\) to \(\pi\): \(0 \leq \phi \leq \pi\) - For the azimuthal angle, we go from \(0\) to \(2\pi\): \(0 \leq \theta \leq 2\pi\)

Computing the Triple Integral

Now, we will compute the triple integral with the correct limits: \(V = \int_{0}^{a} \int_{0}^{\pi} \int_{0}^{2\pi} r^2 \sin{\phi} dr d\phi d\theta\) We'll compute the integral in three steps: 1. Integrate with respect to r 2. Integrate with respect to φ 3. Integrate with respect to θ 1. Integrate with respect to r: \(\int_{0}^{a} r^2 dr = \dfrac{1}{3}r^3 \bigg|_{0}^{a} = \dfrac{1}{3}a^3\) 2. Integrate with respect to φ: \(\int_{0}^{\pi} \sin{\phi} d\phi = -\cos{\phi} \bigg|_{0}^{\pi} = 2\) 3. Integrate with respect to θ: \(\int_{0}^{2\pi} d\theta = \theta \bigg|_{0}^{2\pi} = 2\pi\)

Finding the Volume

Finally, we multiply the results found in the previous steps to obtain the volume of the ball: \(V = \dfrac{1}{3}a^3 \cdot 2 \cdot 2\pi = \dfrac{4}{3}\pi a^3\) So, the volume of a ball with radius a in spherical coordinates is \(\dfrac{4}{3}\pi a^3\).

Key concepts

Triple Integral

A Triple Integral is a fundamental concept in calculus that deals with finding the volume under a surface in a three-dimensional space. When we talk about its application in spherical coordinates, it's primarily used to evaluate volumes of objects with a symmetrical, round shape like a ball or sphere.

A triple integral is expressed as \[ \int \int \int f(x, y, z) \, dx \, dy \, dz \]where the function \(f\) represents the density or some other quantity over the region we're interested in. To find the volume, this function is often simply 1. However, in spherical coordinates, we tweak this expression to\ \[ \int \int \int f(r, \theta, \phi) \, r^2 \sin{\phi} \, dr \, d\phi \, d\theta \]The added term \(r^2 \sin{\phi}\) accounts for the volume element in spherical coordinates.

Setting up the integral involves understanding the limits for \(r\), \(\theta\), and \(\phi\) that describe the volume of the object. To integrate over a full sphere:

• The radius \(r\) goes from the center out to the surface, ranging from \(0\) to \(a\).
• The angle \(\phi\), which is the polar angle, ranges from \(0\) to \(\pi\).
• The angle \(\theta\), which is the azimuthal angle, ranges from \(0\) to \(2\pi\).

Volume Calculation

When calculating the volume of a sphere using the triple integral in spherical coordinates, we set aside our usual rectangular (Cartesian) approach. Instead, spherical coordinates allow us to leverage symmetry, making the calculus much simpler.

The setup is crucial:

• Express the radius \(r\) in terms of its own variable, with limits from the center \(0\) to the surface \(a\).
• Determine the suitable angles \(\phi\) and \(\theta\), which cover the sphere's whole surface.

Imagine slicing the ball like a layered cake, from inside out and top to bottom. The integral \[ V = \int_{0}^{a} \int_{0}^{\pi} \int_{0}^{2\pi} r^2 \sin{\phi} \, dr \, d\phi \, d\theta \]accurately captures the entire volume. When done right, it acknowledges every tiny piece of the sphere, adding them altogether.

Integrating step-by-step in proper order \(dr\), \(d\phi\), and \(d\theta\):

• Begin with \(dr\) to find the radial contribution.
• Follow with \(d\phi\) for angle \(\phi\), capturing polar variation.
• End with \(d\theta\) for azimuthal symmetry around the z-axis.

Ball Volume Formula

The Ball Volume Formula is celebrated because it gives the volume of a sphere with ease, and it is rooted in integral calculus principles. To find the volume of a sphere (or ball) with radius \(a\), the formula derived from our completed triple integral is:\[ V = \frac{4}{3}\pi a^3 \]

In this formula:

• The \(\frac{4}{3}\) factor arises from integrating the radial part.
• The \(\pi\) comes from integrating over the full azimuthal rotation, \(\theta\).
• The \(a^3\) shows the cubic growth of volume as radius increases.

This formula elegantly combines spherical coordinate integration results and common geometric intuition about spheres. What's more, its symmetrical, round nature means whenever you have the radius, simply plug it into the formula for the volume.

Real-world applications range from determining quantities of liquid in spherical tanks to comprehending celestial bodies' sizes. Understanding this formula not only addresses mathematical challenges but also ties into physical fluid dynamics, astrophysics, and volumetric studies.