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# Coordinates (elementary mathematics)

(Redirected from Spherical coordinate system)

This article describes some of the common coordinate systems that appear in elementary mathematics. For advanced topics, please refer to coordinate system. For more background, see Cartesian coordinate system.

The coordinates of a point are the components of a tuple of numbers used to represent the location of the point in the plane or space. A coordinate system is a plane or space where the origin and axes are defined so that coordinates can be measured.

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## Cartesian coordinates

In the two-dimensional Cartesian coordinate system, a point P in the xy-plane is represent by a tuple of two components (x,y).

• x is the signed distance from the y-axis to the point P, and
• y is the signed distance from the x-axis to the point P.

In the three-dimensional Cartesian coordinate system, a point P in the xyz-space is represent by a tuple of three components (x,y,z).

• x is the signed distance from the yz-plane to the point P,
• y is the signed distance from the xz-plane to the point P, and
• z is the signed distance from the xy-plane to the point P.

## Polar coordinates

The polar coordinate systems are coordinate systems in which a point is identified by a distance from some fixed feature in space and one or more subtended angles.

The term polar coordinates often refers to circular coordinates (two-dimensional). Other commonly used polar coordinates are cylindrical coordinates and spherical coordinates (both three-dimensional).

## Circular coordinates

The circular coordinate system, often referred to simply as the polar coordinate system, is a two-dimensional polar coordinate system, defined by an origin, O, and a semi-infinite line L leading from this point. L is also called the polar axis. In terms of the Cartesian coordinate system, one usually picks O to be the origin (0,0) and L to be the positive x-axis (the right half of the x-axis).

In the circular coordinate system, a point P is represented by a tuple of two components (r,θ). Using terms of the Cartesian coordinate system,

• $0\leq{r}$ (radius) is the distance from the origin to the point P, and
• $0\leq\theta<360^\circ$ (azimuth) is the angle between the positive x-axis and the line from the origin to the point P.

## Cylindrical coordinates

The cylindrical coordinate system is a three-dimensional polar coordinate system.

In the cylindrical coordinate system, a point P is represented by a tuple of three components (r,θ,h). Using terms of the Cartesian coordinate system,

• $0\leq{r}$ (radius) is the distance between the z-axis and the point P,
• $0\leq\theta<360^\circ$ (azimuth or longitude) is the angle between the positive x-axis and the line from the origin to the point P projected onto the xy-plane, and
• h (height) is the signed distance from xy-plane to the point P.
Note: some sources use z for h; there is no "right" or "wrong" convention, but it is necessary to be aware of the convention being used.

Cylindrical coordinates involve some redundancy; θ loses its significance if r = 0.

Cylindrical coordinates are useful in analyzing systems that are symmetrical about an axis. For example the infinitely long cylinder that has the Cartesian equation x2 + y2 = c2 has the very simple equation r = c in cylindrical coordinates.

## Spherical coordinates

The spherical coordinate system is a three-dimensional polar coordinate system.

In the spherical coordinate system, a point P is represented by a tuple of three components (ρ,φ,θ). Using terms of the Cartesian coordinate system,

• $0\leq\rho$ (radius) is the distance between the point P and the origin,
• $0\leq\phi\leq 180^\circ$ (colatitude or polar angle) is the angle between the z-axis and the line from the origin to the point P, and
• $0\leq\theta<360^\circ$ (azimuth or longitude) is the angle between the positive x-axis and the line from the origin to the point P projected onto the xy-plane.

NB: The above convention is the standard used by American mathematicians and American calculus textbooks. However, most physicists, engineers, and non-American mathematicians interchange the symbols φ and θ above, using φ to denote the azimuth and θ the colatitude. One should be very careful to note which convention is being used by a particular author. It should be noted that, regardless of how one labels the coordinates, one argument against the conventional American mathematical definition is the fact that it produces a left-handed coordinate system, rather than the usual convention of a right-handed coordinate system.

The spherical coordinate system also involves some redundancy; φ loses its significance if ρ = 0, and θ loses its significance if ρ = 0 or φ = 0 or $\phi=180^\circ$.

To construct a point from its spherical coordinates: from the origin, go ρ along the positive z-axis, rotate φ about y-axis toward the direction of the positive x-axis, and rotate θ about the z-axis toward the direction of the positive y-axis.

Spherical coordinates are useful in analyzing systems that are symmetrical about a point; a sphere that has the Cartesian equation x2 + y2 + z2 = c2 has the very simple equation ρ = c in spherical coordinates.

Spherical coordinates are the natural coordinates for physical situations where there is spherical symmetry. In such a situation, one can describe waves using spherical harmonics. Another application is ergonomic design, where ρ is the arm length of a stationary person and the angles describe the direction of the arm as it reaches out.

The concept of spherical coordinates can be extended to higher dimensional spaces and are then referred to as hyperspherical coordinates.

## Conversion between coordinate systems

### Cartesian and circular

$x=r\,\cos\theta \quad$
$y=r\,\sin\theta \quad$
$r=\sqrt{x^2 + y^2}$
$\theta = \arctan\frac{y}{x} + \pi u_0(-x) \, \operatorname{sgn} y$

where u0 is the Heaviside step function with u0(0) = 0 and sgn is the signum function. Here the u0 and sgn functions are being used as "logical" switches which are used as shorthand substitutes for several if ... then statements. Some computer languages include a bivariate arctangent function atan2(y,x) which finds the value for θ in the correct quadrant given x and y.

### Cartesian and cylindrical

$x=r\,\cos\theta$
$y=r\,\sin\theta$
$z=h \quad$
$r=\sqrt{x^2 + y^2}$
$\theta =\arctan\frac{y}{x} + \pi u_0(-x) \, \operatorname{sgn} y$
$h=z \quad$
$\begin{bmatrix}dx\\dy\\dz\end{bmatrix}= \begin{bmatrix} \cos\theta&-r\sin\theta&0\\ \sin\theta&r\cos\theta&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix}dr\\d\theta\\dh\end{bmatrix}$
$\begin{bmatrix}dr\\d\theta\\dh\end{bmatrix}= \begin{bmatrix} \frac{x}{\sqrt{x^2+y^2}}&\frac{y}{\sqrt{x^2+y^2}}&0\\ \frac{-y}{x^2+y^2}&\frac{x}{x^2+y^2}&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix}dx\\dy\\dz\end{bmatrix}$

### Cartesian and spherical

${x}=\rho \, \sin\phi \, \cos\theta \quad$
${y}=\rho \, \sin\phi \, \sin\theta \quad$
${z}=\rho \, \cos\phi \quad$
${\rho}=\sqrt{x^2+y^2+z^2}$
${\phi}=\arccos\frac{z}{\rho}$
${\phi}=\arctan\frac{\sqrt{x^2+y^2}}{z}$
${\theta} =\arctan\frac{y}{x} + \pi\, u_0(-x)\, \operatorname{sgn} y$
$\begin{bmatrix}dx\\dy\\dz\end{bmatrix}= \begin{bmatrix} \sin\phi\cos\theta&\rho\cos\phi\cos\theta&-\rho\sin\phi\sin\theta\\ \sin\phi\sin\theta&\rho\cos\phi\sin\theta&\rho\sin\phi\cos\theta\\ \cos\phi&-\rho\sin\phi&0 \end{bmatrix} \begin{bmatrix}d\rho\\d\phi\\d\theta\end{bmatrix}$
$\begin{bmatrix}d\rho\\d\phi\\d\theta\end{bmatrix}= \begin{bmatrix} \frac{x}{\rho}&\frac{y}{\rho}&\frac{z}{\rho}\\ \frac{xz}{\rho\sqrt{x^2+y^2}}&\frac{yz}{\rho\sqrt{x^2+y^2}}&\frac{-(x^2+y^2)}{\rho\sqrt{x^2+y^2}}\\ \frac{-y}{\sqrt{x^2+y^2}}&\frac{x}{\sqrt{x^2+y^2}}&0 \end{bmatrix} \begin{bmatrix}dx\\dy\\dz\end{bmatrix}$

### Cylindrical and spherical

${r}=\rho \,\sin\phi$
${\theta}=\theta \quad$
${h}=\rho \,\cos\phi$
${\rho}=\sqrt{r^2+h^2}$
${\phi} =\arctan\frac{h}{r} + \pi \, u_0(-r) \, \operatorname{sgn} h$
${\theta}=\theta \quad$
$\begin{bmatrix}dr\\d\theta\\dh\end{bmatrix}= \begin{bmatrix} \sin\phi&\rho\cos\phi&0\\ 0&0&1\\ \cos\phi&-\rho\sin\phi&0 \end{bmatrix} \begin{bmatrix}d\rho\\d\phi\\d\theta\end{bmatrix}$
$\begin{bmatrix}d\rho\\d\phi\\d\theta\end{bmatrix}= \begin{bmatrix} \frac{r}{\sqrt{r^2+h^2}}&0&\frac{h}{\sqrt{r^2+h^2}}\\ \frac{-h}{r^2+h^2}&0&\frac{r}{r^2+h^2}\\ 0&1&0 \end{bmatrix} \begin{bmatrix}dr\\d\theta\\dh\end{bmatrix}$