Precalculus

Analytic Geometry

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Cylindrical Coordinates

To determine the position of a point in three-dimensional space, cylindrical coordinates are often used. Cylindrical coordinates are introduced as follows. Choose a plane π and define on it a polar coordinate system with a pole O and a polar axis Ox. Cylindrical coordinates extend the polar coordinate system by adding the vertical z-axis passing through the pole O and perpendicular to the plane π.

Cylindrical coordinate system
Figure 1.

Let M be any point in space, N is its orthogonal projection onto the plane π, and \(\left({\rho, \varphi}\right)\) are the polar coordinates of point N on the plane π. The point Mz is the projection of point M on the z-axis so that

\[OM_z = z,\;-\infty \lt z \lt \infty.\]

The ordered triplet of numbers \(\left({\rho, \varphi, z}\right)\) is called the cylindrical coordinates of the point M.

If some point M has coordinates \(\rho,\) \(\varphi,\) \(z\) this is denoted as \(M\left({\rho, \varphi, z}\right).\)

The name "cylindrical coordinates" is due to the fact that the coordinate surface \(\rho = const\) (that is, the surface all the points of which have the same \(\rho-\)coordinate) is a cylinder with the generatrix parallel to the \(z-\)axis.

Converting Between Cylindrical and Cartesian Coordinates

Let the cylindrical and Cartesian coordinate systems have a common origin at point \(O.\) If you choose the axes of the Cartesian coordinate system as indicated in the figure, then the Cartesian coordinates \(\left({x, y, z}\right)\) of the point M will be related to its cylindrical coordinates \(\left({\rho, \varphi, z}\right)\) by the formulas

\[x = \rho\cos\varphi,\;\;y = \rho\sin\varphi,\;\;z = z.\]

The inverse conversion from Cartesian to cylindrical coordinates is given by the expressions

\[\rho = \sqrt{x^2 + y^2},\;\;\tan\varphi = \frac{y}{x},\;\;z=z.\]

Solved Problems

Click or tap a problem to see the solution.

Example 1

Find cylindrical coordinates of the points

  1. \(A\left({-1,-1,1}\right)\)
  2. \(B\left({-3,4,5}\right)\)
  3. \(C\left({0,-2,-1}\right)\)

Example 2

Given the cylindrical coordinates of points find their Cartesian coordinates.

  1. \(K\left({1,\frac{\pi}{4},-1}\right)\)
  2. \(L\left({\sqrt{3},\frac{2\pi}{3},0}\right)\)
  3. \(M\left({4,0,-4}\right)\)

Example 3

Given two points in cylindrical coordinates \(A\left({3,\frac{\pi}{2},4}\right)\) and \(B\left({1,\frac{3\pi}{2},7}\right)\) find the distance between them.

Example 4

Given the cylindrical coordinates of the point \(M\left({\rho,\varphi, z}\right)\) find the angle \(\alpha\) between vector \(\mathbf{OM}\) and the \(x-\)axis.

Example 1.

Find cylindrical coordinates of the points

  1. \(A\left({-1,-1,1}\right)\)
  2. \(B\left({-3,4,5}\right)\)
  3. \(C\left({0,-2,-1}\right)\)

Solution.

Calculate the polar radius \(\rho\) and the polar angle \(\varphi\) using the formulas

\[\rho = \sqrt{x^2 + y^2},\;\;\tan\varphi = \frac{y}{x}.\]

The \(z-\)coordinate (applicate) remains unchanged.

  1. \(A\left({-1,-1,1}\right)\)
    \[\rho = \sqrt{\left({-1}\right)^2 + \left({-1}\right)^2} = \sqrt{2},\;\tan\varphi = \frac{-1}{-1} = 1, \Rightarrow \varphi = \frac{\pi}{4};\]
    \[A\left({-1,-1,1}\right) \mapsto A\left({\sqrt{2},\frac{\pi}{4},1}\right);\]
  2. \(B\left({-3,4,5}\right)\)
    \[\rho = \sqrt{\left({-3}\right)^2 + 4^2} = 5,\;\tan\varphi = \frac{4}{-3} = -\frac{4}{3}, \Rightarrow \varphi = \arctan{\left({-\frac{4}{3}}\right)} = -\arctan\frac{4}{3};\]
    \[B\left({-3,4,5}\right) \mapsto B\left({5,-\arctan\frac{4}{3},5}\right);\]
  3. \(C\left({0,-2,-1}\right)\)
    \[\rho = \sqrt{0^2 + \left({-2}\right)^2} = 2,\;\tan\varphi = \frac{-2}{0} = -\infty, \Rightarrow \varphi = \frac{3\pi}{2};\]
    \[C\left({0,-2,-1}\right) \mapsto C\left({2,\frac{3\pi}{2},-1}\right).\]

Example 2.

Given the cylindrical coordinates of points find their Cartesian coordinates.

  1. \(K\left({1,\frac{\pi}{4},-1}\right)\)
  2. \(L\left({\sqrt{3},\frac{2\pi}{3},0}\right)\)
  3. \(M\left({4,0,-4}\right)\)

Solution.

To convert from cylindrical to Cartesian coordinate system, we use the formulas

\[x = \rho\cos\varphi,\;\;y = \rho\sin\varphi,\;\;z = z.\]
  1. \(K\left({1,\frac{\pi}{4},-1}\right)\)
    \[x = 1\cdot\cos\frac{\pi}{4} = 1\cdot1 = 1,\;y = 1\cdot\sin\frac{\pi}{4} = 1\cdot1 = 1,\;z = -1;\]
    \[K\left({1,\frac{\pi}{4},-1}\right) \mapsto K\left({1,1,-1}\right);\]
  2. \(L\left({\sqrt{3},\frac{2\pi}{3},0}\right)\)
    \[x = \sqrt{3}\cos\frac{2\pi}{3} = \sqrt{3}\left({-\frac{1}{2}}\right)= -\frac{\sqrt{3}}{2},\;y = \sqrt{3}\sin\frac{2\pi}{3} = \sqrt{3}\cdot\frac{\sqrt{3}}{2} = \frac{3}{2},\;z = 0;\]
    \[L\left({\sqrt{3},\frac{2\pi}{3},0}\right) \mapsto L\left({-\frac{\sqrt{3}}{2},\frac{3}{2},0}\right);\]
  3. \(M\left({4,0,-4}\right)\)
    \[x = 4\cos0 = 4\cdot1 = 4,\;y = 4\sin0 = 4\cdot0 = 0,\;z = -4;\]
    \[M\left({4,0,-4}\right) \mapsto M\left({4,0,-4}\right).\]

Example 3.

Given two points in cylindrical coordinates \(A\left({3,\frac{\pi}{2},4}\right)\) and \(B\left({1,\frac{3\pi}{2},7}\right)\) find the distance between them.

Solution.

Let's convert the cylindrical coordinates of the points to Cartesian ones:

\[x_A = \rho\cos\varphi = 3\cos\frac{\pi}{2} = 3\cdot0 = 0,\;\;y_A = \rho\sin\varphi = 3\sin\frac{\pi}{2} = 3\cdot1 = 3,\;\;z_A = 4;\]
\[x_B = \rho\cos\varphi = 1\cos\frac{3\pi}{2} = 1\cdot0 = 0,\;\;y_B = \rho\sin\varphi = 1\sin\frac{3\pi}{2} = 1\cdot\left({-1}\right) = -1,\;\;z_B = 7.\]

Hence points \(A\) and \(B\) have the following coordinates \(x,y,z:\)

\[A\left({0,3,4}\right) \text{ and } B\left({0,-1,7}\right).\]

Now you can easily calculate the distance between these points:

\[d\left({AB}\right) = \sqrt{\left({x_B - x_A}\right)^2 + \left({y_B - y_A}\right)^2 + \left({z_B - z_A}\right)^2} = \sqrt{\left({0 - 0}\right)^2 + \left({-1 - 3}\right)^2 + \left({7 - 4}\right)^2} = \sqrt{0 + 16 + 9} = \sqrt{25} = 5.\]

Example 4.

Given the cylindrical coordinates of the point \(M\left({\rho,\varphi, z}\right)\) find the angle \(\alpha\) between vector \(\mathbf{OM}\) and the \(x-\)axis.

Solution.

In the figure below, the height \(MN\) is perpendicular to the \(xy-\)plane (plane π). The straight line \(NL\) lies in the \(xy-\)plane and is perpendicular to the \(x-\)axis. Therefore, by the theorem of three perpendiculars, the line \(ML\) is also perpendicular to the \(x-\)axis, i.e. triangle \(MLO\) is a right triangle.

The angle alpha between vector OM and the x-axis
Figure 2.

The cosine of the angle \(\alpha = \angle LOM\) in the right triangle \(MLO\) is given by the formula

\[\cos\alpha = \frac{OL}{OM}.\]

Express the leg \(OL\) in terms of known cylindrical coordinates:

\[OL = x = \rho\cos\varphi.\]

According to the Pythagorean theorem,

\[OM = \sqrt{ON^2 + MN^2} = \sqrt{\rho^2 + z^2}.\]

As a result we get

\[\cos\alpha = \frac{OL}{OM} = \frac{\rho\cos\varphi}{\sqrt{\rho^2 + z^2}}.\]