Triple Integrals: Spherical coordinates 2

Pure mathematics is, in its way, the poetry of logical ideas, Albert Einstein.

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Mastering Spherical Coordinates: A Complete Guide

Spherical coordinates are a powerful system used to represent points in three-dimensional space using three parameters: a radial distance $\rho$ (rho), the polar angle $\phi$, and the azimuthal angle $\theta$. In other words, it describes or specifies the position of a point in a three-dimensional space by three numbers $(\rho, \phi, \theta)$. This system is particularly effective for problems involving spherical symmetry, such as those involving spheres or radial distributions. Understanding spherical coordinates provides a powerful tool for solving complex problems in multivariable calculus and physics.

This article uses the standard mathematical calculus convention: $\rho$ for radial distance, $\theta$ for azimuthal angle, and $\phi$ for polar angle. Be aware that many physics textbooks swap the symbols for $\theta$ and $\phi$ (Figure i).

The Geographic Analogy: Longitude, Latitude, and Colatitude

The easiest way to understand spherical coordinates is to imagine the Earth:

• $\rho$ (rho): Think of this as your distance from the center of the Earth.
• $\theta$ (theta): Think of this as Longitude. It measures your rotation around the equator.
• $\phi$ (phi): Think of this as Colatitude. Instead of measuring from the equator, we measure downward from the North Pole.

  Colatitude is the complement of the latitude.

Formal Definitions of the Parameters $(\rho, \theta, \phi)$

• $\rho$ is the radial distance. This is the absolute straight-line distance from the origin to a given point $P$ in space. It ranges from 0 to infinity, $0 \le \rho < \infty$.
• $\phi$ is the polar angle, measured downward from the positive $z$-axis to the radial line $\rho$ (the line connecting the origin to the point $P$). It value increases from $0$ (positive $z$-axis), $\frac{\pi}{2}$ (xy-plane), and $\pi$ (negative $z$-axis), that is, $0 \le \phi \le \pi$ (it stops at the negative $z$-axis, or South Pole).
  It describes how far the point has “dropped” from the North Pole (the $z$-axis).
• $\theta$ is the azimuthal angle, measured counterclockwise from the positive $x$-axis to the projection of the radial line onto the $xy$-plane. It ranges from $0$ to $2\pi$, representing a full rotation around the z-axis. \ It describes the rotation around the z-axis. This is exactly the same $\theta$ used in polar and cylindrical coordinates.

Conversion to Cartesian Coordinates

To convert a point from spherical $(\rho, \theta, \phi)$ to Cartesian $(x, y, z)$, we can use basic trigonometry.

First, look at the vertical triangle formed with the $z$-axis. The vertical height is $z = \rho \cos(\phi)$. The horizontal projection onto the $xy$-plane is $r = \rho \sin\phi$.

By substituting this $r$ into our standard cylindrical conversion formulas ($x = r \cos(\theta)$, $y = r \sin(\theta)$), we get the complete transformation equations: $\begin{aligned} x &= \rho \sin\phi \cos\theta \\ y &= \rho \sin\phi \sin\theta \\ z &= \rho \cos\phi \end{aligned}$

These relationships describe how a point in spherical coordinates maps onto the $x$, $y$, and $z$ axes of Cartesian coordinates. The inverse transformations are given by, $\rho = \sqrt{x^2 + y^2 + z^2}, \phi = \arccos\left(\frac{z}{\sqrt{x^2 + y^2 + z^2}}\right)$, and $\tan(\theta) = \frac{y}{x}$ (adjusted based on the $(x,y)$ quadrant).

Calculating Volume Element dV

When setting up triple integrals in spherical coordinates, you cannot simply use $dV = d\rho \, d\phi \, d\theta$. Because the grid lines curve and spread out the further you get from the origin, the volume of a “differential wedge” changes depending on its location.

Imagine a tiny 3D curved brick in space. Its three side lengths are:

• The change in radial distance: $d\rho$.
• The arc length moving vertically (polar): $\rho \, d\phi$.
• The arc length moving horizontally (azimuthal). Because the horizontal radius is $r = \rho \sin\phi$, this arc length is: $\rho \sin\phi \, d\theta$

Multiplying these three dimensions together yields the proper volume element (which can also be rigorously derived using the Jacobian determinant): $\boxed{dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta}$

Mastering Multiple Integrals: Moments of Inertia, Cylinders, and Gravity

• Evaluate $\int\int\int_{E} (x^2+y^2)dV$ where E is the solid region below the sphere $x^2 + y^2 + z^2 = 1$ and above the $xy$-plane.

1. Physics Note: The expression $(x^2+y^2)$ is the square of the distance from the z-axis. In physics, integrating this over a mass distribution calculates the Moment of Inertia ($I_z$), which measures how resistant the object is to spinning around the z-axis.
2. Because the region $E$ is a hemisphere centered at the origin, spherical coordinates are the perfect tool (Refer to Figure i for a visual representation and aid in understanding it).
   Let’s define our conversions and boundaries: $x = \rho \sin(\phi)\cos(\theta), y = \rho \sin(\phi)\sin(\theta), z = \rho \cos(\phi)$
   $dV = \rho^2\sin(\phi) \, d\rho \, d\phi \, d\theta$
3. The Integrand: We must convert $(x^2+y^2)$ into spherical terms: $x^2+y^2 = (\rho\sin(\phi)\cos(\theta))^2 + (\rho\sin(\phi)\sin(\theta))^2 = \rho^2\sin^2(\phi)(\cos^2(\theta)+\sin^2(\theta))$
   Since $\cos^2(\theta)+\sin^2(\theta) = 1$, the integrand simplifies quite a bit to $\rho^2\sin^2(\phi)$.
4. The Boundaries:
   Radial Distance ($\rho$): From the origin to the sphere’s surface (radius 1). $0 \le \rho \le 1$.
   Polar Angle ($\phi$): From the top of the z-axis ($\phi=0$) down to the xy-plane ($\phi=\frac{\pi}{2}$). $0 \le \phi \le \frac{\pi}{2}$. \ Azimuthal Angle ($\theta$): A full rotation around the z-axis. $0 \le \theta \le 2\pi$
5. Setting Up the Integral: $\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} (\rho^2\sin^2(\phi)) (\rho^2\sin(\phi)) \, d\rho \, d\phi \, d\theta = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \rho^4 \sin^3(\phi) \, d\rho \, d\phi \, d\theta$
6. Evaluate the Integral.
   Integrating with respect to $\rho$: $\int_{0}^{1} \rho^4d\rho = \frac{\rho^5}{5}\bigg|_{0}^{1} = \frac{1}{5}$
   Integrating with respect to $\phi$: $\frac{1}{5} \int_{0}^{\frac{\pi}{2}} \sin^3(\phi) \, d\phi = \frac{1}{5} \int_{0}^{\frac{\pi}{2}} \sin(\phi)(1-\cos^2(\phi)) \, d\phi$
   $= \frac{1}{5} \left( \int_{0}^{\frac{\pi}{2}} \sin(\phi) \, d\phi - \int_{0}^{\frac{\pi}{2}} \sin(\phi)\cos^2(\phi) \, d\phi \right)$
   The first part is obviously quite simple: $\int_{0}^{\frac{\pi}{2}} \sin(\phi) \, d\phi = [-\cos(\phi)]_{0}^{\frac{\pi}{2}} = 0 - (-1) = 1$.
   For the second part, use u-substitution: let $u = \cos(\phi)$, then $du = -\sin(\phi) \, d\phi$. We must also change our limits: when $\phi=0, u=1$, and when $\phi=\frac{\pi}{2}, u=0$.
   $\int_{1}^{0} -u^2 \, du = \int_{0}^{1} u^2 \, du = \left[ \frac{u^3}{3} \right]_{0}^{1} = \frac{1}{3}$ \ Combining both results, we get: $\frac{1}{5} \left( 1 - \frac{1}{3} \right) = \frac{1}{5} \left( \frac{2}{3} \right) = \frac{2}{15}$. \ With respect to $\theta$: $\int_{0}^{2\pi} \frac{2}{15} \, d\theta = \frac{2}{15} [\theta]_{0}^{2\pi} = \frac{4\pi}{15}$

• Set up the triple integrals that represent the volume of a cylinder of radius 2 and height 3 in Cartesian, Cylindrical, and Spherical coordinates (Refer to Figure ii for a visual representation and aid in understanding it).

In rectangular (Cartesian) coordinates, the cylinder is aligned with the z-axis, and its equation is defined by the following constraints: \ The cylinder has a radius of 2, which means that $x^2 + y^2 \le 4$ (this represents the circular cross-section of the cylinder).
The height of the cylinder is 3, which means z ranges from 0 to 3.

Explanation of the limits:

1. x ranges from −2 to 2, which represents the width of the cylinder’s base.
2. For each fixed x, y ranges from $-\sqrt{4-x^2}$ to $\sqrt{4-x^2}$, which is the projection of the circular cross-section of the cylinder onto the xy-plane.
3. z ranges from 0 to 3, representing the height of the cylinder.

Thus, the volume of the cylinder can be calculated using the following triple integral: $V = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{0}^{3} dzdydx$

Cylindrical coordinates $(r, \theta, z)$ are naturally suited for problems involving cylinders. The volume element in cylindrical coordinates is $dV = rdzdrd\theta$, and the equations of the cylinder in these coordinates are:

• The radius of the cylinder is 2, so r ranges from 0 to 2.
• The height of the cylinder is 3, so z ranges from 0 to 3.
• The angle $\theta$ covers the entire circle, so it ranges from 0 to $2\pi$.

The volume of the cylinder can be calculated using the following triple integral in cylindrical coordinates: $\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{3} rdzdrd\theta$

In spherical coordinates $(\rho, \phi, \theta)$, the integral setup for the volume of a cylinder is more complex because spherical coordinates are not inherently suited for cylindrical shapes. The volume element in spherical coordinates is $dV = \rho^2 \sin(\phi)d\rho d\phi d\theta.$

Forcing a cylinder into spherical coordinates is a masterclass in geometry. Imagine standing at the origin and looking up. If you look straight up, your line of sight hits the flat roof of the cylinder ($z=3$). If you lower your gaze, you eventually look at the vertical walls of the cylinder ($r=2$).

Because the “boundary surface” changes depending on your angle, we must split the integral into two parts. The transition happens exactly at the corner where the roof meets the wall. Looking at a 2D cross-section, this forms a right triangle with an opposite side (radius) of $2$, and an adjacent side (height) of $3$. Therefore, the transition angle is $\phi = \arctan(\frac{2}{3})$.

Part 1: Volume Contribution from the Top of the Cylinder

• The roof of the cylinder corresponds to $z = 3$, which in spherical coordinates is $\rho \cos(\phi) = 3$, hence $\rho = 3\sec(\phi)$. Therefore, $\rho$ for the top of the cylinder is bounded by 0 and $3\sec(\phi)$.
• $\phi$ ranges from 0 to $\arctan(\frac{2}{3}) = \arctan(\frac{2}{3})$, which corresponds to the angle at the edge where the top intersects with the side of the cylinder (a right triangle with an opposite side (radius) of $2$, and an adjacent side (height) of $3$).
• $\theta$ ranges from 0 to $2\pi$, covering the full circle around the z-axis.

The integral representing the volume from this part is: $V_{\text{top}} = \int_{0}^{2\pi} \int_{0}^{\arctan(\frac{2}{3})} \int_{0}^{3\sec(\phi)} \rho^2\sin(\phi) \, d\rho \, d\phi \, d\theta$

Part 2: Volume Contribution from the Side of the Cylinder (the Walls)

• The side or wall of the cylinder is defined by $r = 2$, which in spherical coordinates becomes $\sqrt{x^2+y^2} = 2 \implies \sqrt{\rho^2 \sin^2(\phi)\cos^2(\theta) + \rho^2\sin^2(\phi)\sin^2(\theta)} = 2 \implies \sqrt{\rho^2\sin^2(\phi)} = 2 \implies \rho\sin(\phi) = 2 \implies \rho = 2\csc(\phi)$ where the cosecant (csc) of an angle is the reciprocal of the sine of that angle: $\csc(\phi) = \frac{1}{\sin(\phi)}$.
  Therefore, $\rho$ is bounded by $0$ and $2\csc(\phi)$.
• At the side, $\phi$ ranges from $tan^{-1}(\frac{2}{3})$, which corresponds to the angle at the edge of the cylinder to $\pi/2$, which corresponds to the xy-plane.
• $\theta$ again ranges from 0 to $2\pi$, covering the full circle around the z-axis.

$V_{\text{side}} = \int_{0}^{2\pi} \int_{\arctan(\frac{2}{3})}^{\frac{\pi}{2}} \int_{0}^{2\csc(\phi)} \rho^2\sin(\phi) \, d\rho \, d\phi \, d\theta$

The total volume of the cylinder is the sum of these two integrals.

Calculation of Gravitational Force Exerted by a 3D Object

The gravitational force is a fundamental force that attracts any two objects with mass. According to Newton’s law of universal gravitation, the force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula for the gravitational force is: $F = G\frac{m_1m_2}{r^2}$

where:

1. F is the gravitational force acting between the two objects (masses).
2. $m_1$ and $m_2$ are the masses of the two objects.
3. r is the distance between the centers of their masses.
4. G is the gravitational constant.

Objective

The goal is to calculate the gravitational force $\vec{F}$ exerted by an object on a mass m located at the origin.

To calculate the gravitational force exerted by a small differential mass ΔM located at coordinates $(x, y, z)$ on a mass $m$ at the origin, we use the formula:

$\vert \vec{F} \vert = G\frac{\Delta M \cdot m}{\rho^2},\text{ and the direction is given by } dir(\vec{F}) = \frac{\langle x, y, z \rangle}{\rho}$ (the direction of the pull is the unit vector pointing from the origin to the tiny chunk of mass $\Delta M$) where:

1. $\vert \vec{F} \vert$ is the magnitude of the gravitational force.
2. $\rho$ is the distance from the differential mass to the origin: $\rho = \sqrt{x^2+y^2+z^2}$
3. $\langle x, y, z \rangle$ is the position vector of the differential mass.

Multiplying the magnitude by the direction vector gives the vector force: $\vec{F} = \frac{G \cdot \Delta M \cdot m}{\rho^3}\langle x, y, z \rangle$ (Figure C).

To find the gravitational force exerted by the entire object, we integrate this gravitational force over the entire volume occupied by the object.

Given that $\Delta M = \delta \cdot \Delta V$, where $\delta$ is the density and $\Delta V$ is the volume element, we can express the total gravitational force as: $\vec{F} = \int \int \int \frac{G·m·⟨x, y, z⟩}{\rho^3}δdV$

This integral can be complex, so it is useful to simplify by using spherical coordinates. In spherical coordinates, $z = \rho\cos(\phi)$, and the volume element is $dV = \rho^2 \sin(\phi) d\rho d\phi d\theta$. If we place the solid in a way that the z-axis is an axis of symmetry, the gravitational force will be directed along the z-axis (the x and y components will perfectly cancel each other out) -Figure D-, that is, it will reduce the force vector to $\vec{F} = \langle 0, 0, F_z \rangle$. In other words, the net force only pulls straight up or down along the z-axis.

Therefore, the integral simplifies to: $F_z = \int \int \int \frac{G \cdot m \cdot z}{\rho^3}\delta dV$

Let’s switch to spherical coordinates! We know $z = \rho\cos(\phi)$ and $dV = \rho^2\sin(\phi) \, d\rho \, d\phi \, d\theta$

$F_z = G m \int \int \int \frac{\rho \cos(\phi)}{\rho^3}\delta \cdot \rho^2\sin(\phi) d\rho d\phi d\theta = G m \int \int \int \delta \cdot \cos(\phi) \sin(\phi)d\rho d\phi d\theta.$

Notice how the $\rho^3$ in the denominator is perfectly canceled by the $\rho \cdot \rho^2$ in the numerator!

Application: Force of a Solid Hemisphere

Let’s apply this beautifully simplified formula to calculate the force exerted by a uniform solid hemisphere of radius $R$ resting on the xy-plane above the mass $m$.

Limits: $0 \le \rho \le R$, $0 \le \phi \le \frac{\pi}{2}$, and $0 \le \theta \le 2\pi$.

$F_z = Gm\delta \int_{0}^{2\pi} d\theta \int_{0}^{\frac{\pi}{2}} \cos(\phi)\sin(\phi) \, d\phi \int_{0}^{R} d\rho$

Evaluating each piece independently:

1. $\int_{0}^{R} d\rho = R$
2. $\int_{0}^{\frac{\pi}{2}} \cos(\phi)\sin(\phi) \, d\phi = \left[ \frac{\sin^2(\phi)}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2}$
3. $\int_{0}^{2\pi} d\theta = 2\pi$

Multiplying them together: $F_z = Gm\delta (2\pi) \left(\frac{1}{2}\right) (R) = \pi G m \delta R$

We can make this even more elegant. The total mass $M$ of a hemisphere is its volume times its density: $M = \frac{2}{3}\pi R^3 \delta$. Solving for density gives $\delta = \frac{3M}{2\pi R^3}$.

Substituting this into our force equation: $F_z = \pi G m \left( \frac{3M}{2\pi R^3} \right) R = \frac{3GmM}{2R^2}$

This final equation is incredible. It reveals that the gravitational pull of a solid hemisphere on a point at its base is exactly $1.5$ times the pull if all that mass were simply concentrated into a point mass at a distance $R$. Spherical coordinates made this advanced physics proof a walk in the park!

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