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.

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 M_z is the projection of point M on the z-axis so that

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

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

Solved Problems

Solution.

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

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

Solution.

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

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. \(M\left({4,0,-4}\right)\)
   \[x = 4\cos0 = 4\cdot1 = 4,\;y = 4\sin0 = 4\cdot0 = 0,\;z = -4;\]
   \[M\left({4,0,-4}\right) \mapsto M\left({4,0,-4}\right).\]

Solution.

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

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

Now you can easily calculate the distance between these points:

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

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

According to the Pythagorean theorem,

As a result we get