#24 Converting P(−2, 6, 3) to cylindrical and spherical coordinates

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. The given Cartesian point

   

   Identify the given point before changing coordinates.

   This worked example concerns a point in three-dimensional Euclidean space.

   A coordinate conversion describes the same point with a different ordered triple.

   Cartesian components:

   $\displaystyle x = -2, y = 6, z = 3$

   Given Cartesian point:

   $\displaystyle P = \left(-2, 6, 3\right)$

   Find the cylindrical triple (ρ, φ, z) and spherical triple (r, θ, φ).

   Narration transcript

   Hi everyone. This is the first worked example of our Electromagnetic Theory series. We are going to do a coordinate conversion. We have a point P in Cartesian coordinates: x equals negative two, y equals six, z equals three. So P is at negative two, six, three. We need to express this same point first in cylindrical coordinates, then in spherical coordinates.

2. 2. Cylindrical coordinates and quadrant

   

   Begin with the cylindrical representation.

   The coordinate order is radial distance ρ, azimuth φ, and height z.

   The cylindrical radius is the perpendicular distance from the z-axis.

   Measure azimuth counterclockwise from the positive x-axis in the xy-plane.

   The height remains the Cartesian z-coordinate.

   Cylindrical radius:

   $\displaystyle \rho = \sqrt{x^{2}+y^{2}}$

   Principal inverse-tangent value before quadrant correction:

   $\displaystyle \alpha = \arctan \left(\frac{y}{x}\right)$

   The unchanged height is the third cylindrical coordinate.

   Substitute the Cartesian values:

   $\displaystyle x = -2, y = 6$

   Keep the negative coordinate inside parentheses when squaring:

   $\displaystyle \rho = \sqrt{\left(-2\right)^{2}+6^{2}} = \sqrt{4+36} = \sqrt{40}$

   Exact radius and decimal approximation:

   $\displaystyle \rho = 2\sqrt{10} \approx 6.325$

   Positive y and negative x place the point in quadrant II.

   The calculator returns the principal value in quadrant IV:

   $\displaystyle \alpha = \arctan \left(\frac{6}{-2}\right) \approx -71.57^{\circ}$

   Add 180 degrees to select quadrant II:

   $\displaystyle \varphi = 180^{\circ}-71.57^{\circ} = 108.43^{\circ}$

   The height is unchanged:

   $\displaystyle z = 3$

   Cylindrical coordinates, with rounded azimuth:

   $\displaystyle P = \left(2\sqrt{10}, 108.43^{\circ}, 3\right)$

   Narration transcript

   Let's start with cylindrical. In cylindrical coordinates a point is described by three values: rho, phi, and z. rho is the perpendicular distance from the z-axis. phi is the angle measured from the positive x-axis in the counter-clockwise direction. And z stays exactly the same z we had in Cartesian. The formulas are: rho equals the square root of x squared plus y squared. phi equals the arctangent of y over x, but be careful, you must pick the correct quadrant. And z equals z. Let's plug in: x is negative two, y is six. rho equals the square root of negative two squared, which is four, plus six squared, which is thirty six; together forty inside the root. Square root of forty is two root ten, approximately six point three two five. Now phi: y is positive and x is negative, so our point sits in the second quadrant. A calculator gives arctangent of six over negative two as about minus seventy one point five seven degrees, but that is a fourth-quadrant answer. To land in the second quadrant, we add one hundred eighty degrees: one hundred eighty minus seventy one point five seven equals one hundred eight point four three degrees. z is still three. So in cylindrical, P is: rho equals two root ten, phi equals one hundred eight point four three degrees, z equals three.

3. 3. Spherical radius and polar angle

   

   Now express the same point in spherical coordinates.

   Use the ordered triple (r, θ, φ).

   The radius r is the distance from the origin; θ is the polar angle down from +z; φ is the same xy-plane azimuth.

   Spherical radius:

   $\displaystyle r = \sqrt{x^{2}+y^{2}+z^{2}}$

   Polar angle measured from the positive z-axis:

   $\displaystyle \theta = \arccos \left(\frac{z}{r}\right)$

   The azimuth is shared by the cylindrical and spherical descriptions.

   Compute the radius:

   $\displaystyle r = \sqrt{40+9} = \sqrt{49} = 7$

   Compute the polar angle:

   $\displaystyle \theta = \arccos \left(\frac{3}{7}\right) \approx 64.62^{\circ}$

   Reuse the quadrant-corrected azimuth:

   $\displaystyle \varphi = 108.43^{\circ}$

   Spherical coordinates, with rounded angles:

   $\displaystyle P = \left(7, 64.62^{\circ}, 108.43^{\circ}\right)$

   Narration transcript

   Now spherical. Again three values: r, theta, and phi. Here r is the distance from the origin straight to the point; theta is the angle measured down from the positive z-axis, called the polar angle; and phi is the same azimuthal angle we just computed. Formulas: r equals the square root of x squared plus y squared plus z squared. theta equals the arccosine of z over r. phi is identical to the cylindrical phi. Let's compute: r equals the square root of forty plus nine, square root of forty nine, exactly seven. theta equals arccosine of three over seven, which is about sixty four point six two degrees. phi we already know: one hundred eight point four three degrees. So in spherical, P is: r equals seven, theta equals sixty four point six two degrees, phi equals one hundred eight point four three degrees.

4. 4. Geometric consistency

   

   The numerical conversion is complete.

   Check the signs and angles against the geometric position.

   The signs are negative x, positive y, and positive z.

   The xy projection is in quadrant II and the point is above the xy-plane.

   An azimuth near 108 degrees lies between 90 and 180 degrees, as required.

   A polar angle near 65 degrees lies between the positive z-axis and the xy-plane; it is neither a pole nor an equatorial point.

   The straight distance from the origin is seven units:

   $\displaystyle r = 7$

   The two plane views and the coordinate calculations are consistent.

   Narration transcript

   We have the numbers. Now a quick geometric sanity check. Our point has x negative, y positive, z positive. So in the xy-plane it lies in the second quadrant, and it sits above the xy-plane. phi around one hundred degrees is consistent with being in the second quadrant. theta around sixty five degrees means the point is at a moderate tilt away from the z-axis, not at the pole, not at the equator. And r equals seven, a straight line of seven units from the origin. Everything matches the picture.

5. 5. Coordinate triples recap

   

   Recap the coordinate order and the role of each angle.

   The Cartesian point (-2, 6, 3) becomes the cylindrical triple (2√10, 108.43°, 3).

   The spherical triple is (7, 64.62°, 108.43°), with θ measured from +z.

   The next topic concerns vector fields in different coordinate systems; transforming a point is distinct from transforming vector components.

   Keep the point fixed and check the quadrant whenever changing its coordinates.

   Narration transcript

   Quick recap. The point that was negative two, six, three in Cartesian becomes two root ten, one hundred eight point four three degrees, three in cylindrical. And seven, sixty four point six two degrees, one hundred eight point four three degrees in spherical. In the next video we will see how vector fields transform between these three systems. See you then.

Source video: Electromagnetic Theory (v2) #24 | Problem Solving #01: Coordinate Conversion (4:26)