Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When learning vector fields and using Green's Theorem with the Jacobian to find the area of a level surface, I actually realized that most of the examples shown in my book would be much easier to solve by using polar and spherical coordinates, but not multiplying by the representation of the radius. (Eg: In spherical coordinates I would eliminate the $\rho^2 \sin(\phi)$. I ended up discussing with my professor about it, and he said that the radius should be fixed. I disagreed, saying that the radius is changing, but you are not multiplying by it.

Here is how I see the use of Spherical Coordinates for getting the volume:

My idea is that, in spherical coordinates, if you take $z = \rho \cos(\theta)$, and look at the Cartesian plane from the z-axis, you are actually adding all the circles on the xy-plane. Then by looking from the y-axis, you have other sets of circles, and that's where you are getting your radius for multiplying $\rho^2 \sin(\phi)$. Is this right or I understood it wrongly?

Thanks for your attention

share|cite|improve this question
up vote 2 down vote accepted

I'm not sure exactly what you're asking, but whenever you have a change of coordinates, you get some "stretching." For instance, with sphereical coordinates, you're using three variables $\rho, \theta, \phi$ (which you can think of as piece of regular old $\mathbb{R}^3$) to describe some "curved" space. For instance, the regular old box $[1,2]\times[0,\pi/2]\times[0,\pi/2]$ describes the spherical regionin the first octant between the radii 1 and 2. These two geometric shapes do not have the same volume! you have to "stretch" the box to get the spherical shape, and the $\rho^2\sin\phi$ measures that stretching. In general if you have a change of coordinates like $$x(\rho, \theta, \phi)=\rho\sin\phi\cos\theta,y(\rho, \theta, \phi)=\rho\sin\phi\sin\theta,z(\rho, \theta, \phi)=\rho\cos\phi $$ you calculate the stretching by computing the the determinant $$ \left| \begin{array}{ccc} \partial x/\partial\rho&\partial x/\partial\theta&\partial x/\partial\phi\\ \partial y/\partial\rho&\partial y/\partial\theta&\partial y/\partial\phi\\ \partial z/\partial\rho&\partial z/\partial\theta&\partial z/\partial\phi\\ \end{array} \right|=\rho^2\sin\phi $$ if you're familar with the notion of the determinant describing the volume of a parallepiped, then you can think of this procedure as seeing how a small box near $(\rho,\theta,\phi)$ spaces changes shape into a small parallelepiped $(x,y,z)$ (using the partial derivatives as a linear approximation to your coordinate change). most calculus books will have a discussion of this in the multiple integration part of the book.

share|cite|improve this answer
I think I got it. My mistake was to assume that p^2 sin(phi) represents the radius of the sphere. How would you get the surface area a sphere centered at the origin with radius 1 using spherical coordinates? Setting z = sqrt(-x^2 + -y^2) and using spherical coordinates as parameters would work? – OverAchiever Dec 8 '11 at 19:56
For surface area you would set $\rho = \text{const}$ and integrate over $\theta$ and $\phi$... – David Z Dec 9 '11 at 0:39
Thanks. The explanation on how p^2 sin(phi) represents the stretching really helped. – OverAchiever Dec 10 '11 at 0:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.