Cartesian -> Spherical polar

In summary, when transforming from cartesian coordinates to spherical polar coordinates, the surface element dS becomes dS = r^2 sin(theta) dtheta dphi in the case of a surface defined by x^2 + y^2 + z^2 = 1. This can be understood geometrically by considering an infinitesimal rectangle on the sphere and calculating its area, or can be derived more rigorously using the fundamental vector product. It is worth noting that the notation for theta and phi may vary between mathematics and engineering.
  • #1
mcfc
17
0
I have an integral [tex]\int \int_S x^2 + yz \ dS[/tex]

and wish to transform to spherical polar coordinates. How does dS become

[tex] dS = r^2 \sin \theta d\theta d\phi [/tex]??

Where surface S is [tex]x^2 + y^2 + z^2 = 1[/tex]
 
Physics news on Phys.org
  • #2
There are two ways of looking at this.

First, a rather rough "geometrical" look. Imagine an infinitesmal "rectangle" on the sphere having "top" at [itex]\phi[/itex] and "bottom" at [itex]\phi+ d\phi[/itex], "left" at [itex]\theta[/itex] and "right" at [itex]\theta+ d\theta[/itex]. The left and right, "lines of longitude", are great circles, with radius the radius of the sphere, 1. The length of those two sides is [itex]1(d\phi)= d\phi[/itex]. The top and bottom, "lines of latitude", are not great circles. Their centers lie on the line through the poles, (0,0,1) and (0,0,-1). Given a point on one of those circles, drop a perpendicular to that vertical line. You have a right triangle with one leg being the radius of the circle, r, the hypotenuse the radius of the circle, 1, and angle at the center [itex]\phi[/itex]. Then [itex]r/1= sin(\phi)[/itex] so the radius of the circle is [itex]sin(\phi)[/itex] and the length of the arc on the sphere is [itex]sin(\phi)d\theta[/itex]. The area of the "rectangle" is (width times height) [itex]sin(\phi)d\theta d\phi[/itex].

A more rigorous calculation involves the "fundamental vector product" which is worth knowing on its own. Any smooth surface, since a surface is two dimensional, can be written in terms of two parameters: x= f(u,v), y= g(u,v), z= h(u,v) which can also be written as the vector equation: [itex]\vec{r}= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}[/itex]. The derivatives of that with respect to the two variables, [itex]\vec{r_u}= f_u(u,v)\vec{i}+ g_u(u,v)\vec{j}+ h_u(u,v)\vec{k}[/itex] and [itex]\vec{r_v}= f_v(u,v)\vec{i}+ g_v(u,v)\vec{j}+ h_v(u,v)\vec{k}[/itex] are vectors in the tangent plane to the surface at each point. The cross product then, [itex]\vec{r_u}\times\vec{r_v}[/itex], the "fundamental vector product", is perpendicular to the surface and, because of the derivatives, its length measures the area of an infinitesmal region: [itex]dS= \left|\vec{r_u}\times\vec{r_v}\right|dudv[/itex].

The spherical coordinates are connected to the cartesian coordinates by [itex]x= \rho cos(\theta)sin(\phi)[/itex], [itex]y= \rho sin(\theta)sin(\phi)[/itex], [itex]z= \rho cos(\phi)[/itex] and we can take those as parametric coordinates for the unit sphere by taking [itex]\rho= 1[/itex]: [itex]x= cos(\theta)sin(\phi)[/itex], [itex]y= sin(\theta)sin(\phi)[/itex], [itex]z= cos(\phi)[/itex] or [itex]\vec{r}= cos(\theta)sin(\phi)\vec{i}+ sin(\theta)sin(\phi)\vec{j}+ cos(\phi)\vec{k}[/itex]. The derivatives are [itex]\vec{r_\theta}= -sin(\theta)sin(\phi)\vec{i}+ cos(\theta)sin(\phi)\vec{j}[/itex] and [itex]\vec{r_\phi}= cos(\theta)cos(\phi)\vec{i}+ sin(\theta)cos(\phi)\vec{j}- sin(\phi)\vec{k}[/itex]. The cross product of those is [itex]cos(\theta)sin^2(\phi)\vec{i}+ sin(\theta)sin^2(\phi)\vec{j}+ sin(\phi)cos(\phi)\vec{k}[/itex] which has length [itex]\sqrt{cos^2(\theta)sin^4(\phi)+ sin^2(\theta)sin^4(\phi)+ sin^2(\phi)cos^2(\phi)}[/itex][itex]= \sqrt{sin^4(\phi)+ sin^2(\phi)cos^2(\phi)}[/itex][itex]= \sqrt{sin^2(\phi)(sin^2(\phi)+ cos^2(\phi))}[/itex][itex]= sin(\phi)[/itex]. That gives [itex]dS= sin(\phi)d\phi d\theta[/itex].

By the way, my [itex]\theta[/itex] and [itex]\phi[/itex] are reversed from yours. Mathematics notation (mine) uses [itex]\theta[/itex] as "longitude" and [itex]\phi[/itex] as "co-latitude". Engineering notation reverses those.
 
Last edited by a moderator:

1. What is the difference between Cartesian and Spherical polar coordinates?

Cartesian coordinates are a system of locating points in 3D space using x, y, and z coordinates. Spherical polar coordinates use a different system, using a radius, angle from the z-axis (also known as polar angle), and angle from the x-axis (also known as azimuth angle) to locate points in 3D space.

2. How do you convert from Cartesian to Spherical polar coordinates?

To convert from Cartesian to Spherical polar coordinates, you can use the following formulas:

r = √(x² + y² + z²)

θ = arccos(z/√(x² + y² + z²))

φ = arctan(y/x)

Where r is the radius, θ is the polar angle, and φ is the azimuth angle.

3. Can you convert from Spherical polar to Cartesian coordinates?

Yes, you can convert from Spherical polar to Cartesian coordinates using the following formulas:

x = r sin(θ) cos(φ)

y = r sin(θ) sin(φ)

z = r cos(θ)

4. What are the applications of Spherical polar coordinates?

Spherical polar coordinates are commonly used in physics and engineering, particularly in fields such as electromagnetics, fluid mechanics, and astronomy. They are also useful for describing the position and orientation of objects in 3D space.

5. Are there any limitations to using Spherical polar coordinates?

While Spherical polar coordinates have many applications, they do have some limitations. They are not as intuitive as Cartesian coordinates, and some operations such as addition and subtraction can be more complicated. Additionally, they are not as useful for describing points close to the origin or along the z-axis, where the polar angle becomes undefined.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
488
  • Calculus and Beyond Homework Help
Replies
7
Views
881
  • Calculus and Beyond Homework Help
Replies
4
Views
954
  • Calculus and Beyond Homework Help
Replies
4
Views
920
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
968
Replies
33
Views
3K
  • Calculus and Beyond Homework Help
Replies
7
Views
663
  • Calculus and Beyond Homework Help
Replies
2
Views
145
  • Calculus and Beyond Homework Help
Replies
1
Views
561
Back
Top