Volume element in Spherical Coordinates

[LagrangeEuler]

For me is not to easy to understand volume element $dV$ in different coordinates. In Deckart coordinates $dV=dxdydz$. In spherical coordinates it is $dV=r^2drd\theta d\varphi$. If we have sphere $V=\frac{4}{3}r^3 \pi$ why then
$$dV=4\pi r^2dr$$
always?

 

[S.G. Janssens]

The coordinates are named after Descartes and are usually called "Cartesian coordinates".
I don't think your expression for the spherical volume element is correct: It misses a factor $\sin{\phi}$, so: $dV = \rho^2\sin{\phi}\,d\rho\,d\theta\,d\phi$.

As to the "why": There are various less and more rigorous ways to see it. The rigorous answer is that the factor $\rho^2\sin{\phi}$ arises (up to a sign) as the determinant of the matrix $\frac{\partial (x,y,z)}{\partial (\rho,\theta,\phi)}$ that describes the changes-of-variables from Cartesian to spherical. This is discussed in multivariable calculus books such as Marsden and Tromba's Vector Calculus.

 

[LagrangeEuler]

Yes I forgot $\sin \theta$.

 

[vanhees71]

Yes, look for "Jacobi determinant".

It's also very easy to understand when you remember that for three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ the volume of the parallelipiped spanned by them is $V=|\vec{a} \cdot (\vec{b} \times \vec{c})$. It's easy to understand when you remember the meaning of $\vec{b} \times \vec{c}$ and the fact that the volume of the parallelipiped is given by "area times height".

Now apply this to the given coordinates. They are defined by
$$\vec{x}=\vec{x}(r,\vartheta,\varphi)=\begin{pmatrix} r \sin \vartheta \cos \varphi \\ r \sin \vartheta \sin \varphi \\ r \cos \vartheta \end{pmatrix}.$$
Now look at an arbitrary point at the three "coordinate lines", i.e., the lines which are defined by varying one of the coordinates and leave the other constant. At the point defined by $(r,\vartheta,\varphi)$ you get three tangent vectors spanning a little box-like volume, and this you can use as the volume elements in volume integrals. So what you get is
$$\mathrm{d}^3 x = \mathrm{d} r \mathrm{d} \vartheta \mathrm{d} \varphi |(\partial_r \vec{x}) \cdot (\partial_{\vartheta} \vec{x} \times \partial_{\varphi} \vec{x})|.$$
It's a good exercise to do this calculation and prove that indeed
$$\mathrm{d}^3 x = \mathrm{d} r \mathrm{d} \vartheta \mathrm{d} \varphi r^2 \sin \vartheta.$$
Another way to see it is also to simply draw the coordinate lines and the little box spanned by it. Then you can read off the volume using simple geometry.

 

[etropy]

Are you leaking intuition, or do you have trouble to understand how to derive it?

 

[romsofia]

Another way to derive these area elements is the use of exterior derivatives, which i'll give you a quick example with polar coords. Let $x=r \cos \theta$ and $y = r \sin \theta$. An area element in cartesian coords is simply $dxdy$(which *technically* is $dx \wedge dy$) so, all we have to do is find out what "$dxdy$" is in polar coords! Thus, $dx = \cos \theta dr - r \sin \theta d \theta \text{ and } dy = \sin \theta dr + r \cos \theta d \theta$ From here we see that $$dx \wedge dy = \cos \theta \sin \theta dr \wedge dr +r \cos^2 \theta dr \wedge d \theta - r \sin^2 \theta d \theta \wedge dr -r^2 \sin \theta \cos \theta d \theta \wedge d \theta $$

A few things you may not know is that $\wedge$ is called a "wedge product", $dx \wedge dy = -dy \wedge dx$ (this is what is known as anti-symmetry!) and that $d(anything) \wedge d(anything) = 0$ (this follows from the above property, you may ponder why!). So, cleaning up our formula we see that...
$$dx \wedge dy = r \cos^2 \theta dr \wedge d \theta + r \sin^2 \theta dr \wedge d \theta = r dr \wedge d \theta (\cos^2 \theta + \sin^2 \theta) = r dr \wedge d \theta$$ or when you take the wedges out (since they are just a product...) you get $$dxdy = rdrd\theta$$ The same can be down for spherical coords to find the volume element!