Express the plane z=x in cylindrical and spherical coordinates.

Express The Plane

This question aims to find the cylindrical and spherical coordinates of the plane z = x.

This question is based on the concept of coordinate systems from calculus. Cylindrical and spherical coordinate systems are expressed in the cartesian coordinate systems. A spherical object like a sphere of a ball is best expressed in a spherical coordinate system while cylindrical objects like pipes are best described in the cylindrical coordinate system.

Expert Answer:

The plane z =x is a plane that lies in the xz-plane in the cartesian coordinate system. The graph of plane z=x is shown in Figure 1 and it can be seen that the y-component of the graph is zero.

Cartesian Coordinate System

Figure-1 : Cartesian Coordinate System

We can express this plane in spherical and cylindrical coordinates using their derived formulas.

1) Cylindrical Coordinates are given by:

\[ (x, y, z) = (r \cos \theta, r \sin \theta, z) \quad 0 \leq \theta \leq 2\pi \]


\[ r = \sqrt{x^2 + y^2} \quad r \geq 0 \]

Cylindrical Coordinate System

Figure-2 : Cylindrical Coordinate System


\[ z = x \]

So the equation becomes,

\[ (x, y, z) = (r \cos \theta, r \sin \theta, r \cos \theta) \]

2) Spherical Coordinates are given by:

\[ (x, y, z) = (\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \quad \rho \geq 0,  0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi \]

Spherical Coordinate System

Figure-3 : Spherical Coordinate System


\[ z = x \]

\[ \rho \cos \phi = \rho \sin \phi \cos \theta \]

\[ \dfrac{\cos \phi}{\sin \phi} = \cos \theta \]

\[ \cot \phi = \cos \theta \]

\[ \theta = \arccos (\cot \phi) \]

By substituting the values we get,

\[ (x, y, z) = (\rho \sin \phi \cos (\arccos (\cot \phi)), \rho \sin \phi \sin (\arccos (\cot \phi)), \rho \cos \phi) \]

Simplifying by using trigonometric identities, we get:

\[ (x, y, z) = (\rho \cos \phi, \rho \sin \phi \sqrt{1 – \cot^{2} \phi}, \rho \cos \phi) \]

Numerical Results:

Cylindrical Coordinates,

\[ (x, y, z) = (r \cos \theta, r \sin \theta, r \cos \theta) \]

Spherical Coordinates,

\[ (x, y, z) = (\rho \cos \phi, \rho \sin \phi \sqrt{1 – \cot^{2} \phi}, \rho \cos \phi) \]


Convert $(5, 2, 3)$ cartesian coordinates into cylindrical and spherical coordinates.

Cylindrical Coordinates are given by,

\[ (x, y, z) = (r \cos \theta, r \sin \theta, z) \]


\[ r =5.38 \]


\[ \theta = 21.8^{\circ} \]

By substituting the values, we get,

\[ (x, y, z) = (20.2, 8.09, 3) \]

Spherical Coordinates are given by,

\[ (x, y, z) = (\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \]

We calculated the values of $r$ and $\theta$ above and now we calculate $\rho$ and $\phi$ for spherical coordinates.

\[ \rho = r^2 + z^2 \]

\[ \rho = 6.16 \]

We know that $\phi$ is the angle between $\rho$ and $z-axis$, and by using geometry we know that $\phi$ is also the angle between $\rho$ and the vertical side of the right-angled triangle.

\[ \phi = 90^{\circ} – \theta \]

\[ \phi = 68.2^{\circ} \]

By substituting the values and implying, we get:

\[ (x, y, z) = (5.31, 2.12, 2.28) \]

Images/Mathematical drawings are created with Geogebra.

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