## 17Calculus - Spherical Coordinates

This page covers the basics of spherical coordinates and reference equations that use spherical coordinates for multi-variable calculus.

Spherical coordinates are very different from rectangular and cylindrical coordinates. There are still three coordinates, usually labeled $$(\rho, \theta, \phi)$$, but they are assigned so that spherical-type objects are easy to express. All distances and angles are measured based on a radial line drawn from the origin to the point. Table 1 describes how they are defined and Figure 1 shows this graphically.

Table 1 $$\rho$$ is the radial distance out from the origin to the point $$\theta$$ is the angle measured in the xy-plane from the positive x-axis to the shadow of the radial line. Note: This is the same angle as $$\theta$$ in cylindrical coordinates. $$\phi$$ is the angle measured from the positive z-axis to the radial line
Figure 1 Note - We have specifically chosen $$(\rho, \theta, \phi)$$ to describe spherical coordinates but not all books, instructors and videos use these same variables. Some use $$(r,\theta,\phi)$$ with $$\theta$$ and $$\phi$$ meaning different things depending on the context. (See the Wiktionary page on spherical coordinates for examples.) Check with your instructor and textbook to see which one they require.

$$(x,y,z) \to (\rho, \theta, \phi)$$

$$x=\rho\sin\phi\cos\theta$$

$$y=\rho\sin\phi\sin\theta$$

$$z=\rho\cos\phi$$

$$\rho^2=x^2+y^2+z^2$$

Table 2

The equations to convert from rectangular to spherical coordinates, in Table 2, are somewhat complicated. So take a few minutes to learn and memorize them (why memorize?). Look for similarities and differences between the equations and see if you can make some sense out of them.

Here is a quick video clip discussing these equations and showing a neat animation that will help you get a feel for spherical coordinates.

### MIP4U - Introduction to Spherical Coordinates [2mins-13secs]

video by MIP4U

Converting between Rectangular and Spherical Coordinates

Rectangular → Spherical

In order to convert from rectangular to spherical coordinates, the equations are Table 2 above are directly applied. The technique is not hard but it will help to watch a couple of quick videos showing how to do this with examples. The first video explains how to convert points from rectangular to spherical. The second video explains how to convert equations from rectangular to spherical. Both videos have plenty of examples.

### Krista King Math - Spherical Coordinates [3mins-53secs]

video by Krista King Math

### MIP4U - Converting Between Spherical and Rectangular Equations [8mins-15secs]

video by MIP4U

To be able to use spherical coordinates in triple integrals, you need to understand the unit vectors in spherical coordinates. Here is a video that will help.

### Michel vanBiezen - Spherical Unit Vector Conversions [9mins-45secs]

video by Michel vanBiezen

Before we go on, let's do a few practice problems.

Practice

Find an equation in spherical coordinates for the rectangular equation $$x^2 + y^2 - z^2 = 0$$.

Problem Statement

Find an equation in spherical coordinates for the rectangular equation $$x^2 + y^2 - z^2 = 0$$.

Solution

### 2209 video

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Spherical → Rectangular

Sometimes it will be easier to evaluate an integral in rectangular coordinates rather than spherical coordinates. In this case, you may be given an equation in spherical coordinates and be asked to convert to rectangular coordinates.

To do this conversion, there is no set of equations with specific techniques. You just have to use trig identities and intuition to get an equation in rectangular coordinates. However, it sometimes helps to convert tangent, cotangent, secant and cosecant to sines and cosines and use $$\cos^2(t) + \sin^2(t) = 1$$. Try these practice problems to get the idea.

Practice

Unless otherwise instructed, convert these equations in spherical coordinates to rectangular coordinates.

$$\rho = 2 \csc(\phi)$$

Problem Statement

Convert $$\rho = 2 \csc(\phi)$$ to rectangular coordinates.

$$x^2 + y^2 = 4$$

Problem Statement

Convert $$\rho = 2 \csc(\phi)$$ to rectangular coordinates.

Solution

### 2205 video

$$x^2 + y^2 = 4$$

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You CAN Ace Calculus

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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