Deriving the spherical volume element

• I
• quickAndLucky
In summary, the author is trying to derive the infinitesimal volume element in spherical coordinates, but is having difficulty doing so.
quickAndLucky
I’m trying to derive the infinitesimal volume element in spherical coordinates. Obviously there are several ways to do this. The way I was attempting it was to start with the cartesian volume element, dxdydz, and transform it using

$$dxdydz = \left (\frac{\partial x}{\partial r}dr + \frac{\partial x}{\partial \theta }d\theta + \frac{\partial x}{\partial \phi }d\phi \right )\left ( \frac{\partial y}{\partial r}dr + \frac{\partial y}{\partial \theta }d\theta + \frac{\partial y }{\partial \phi}d\phi \right )\left ( \frac{\partial z}{\partial r}dr + \frac{\partial z}{\partial \theta }d\theta + \frac{\partial z}{\partial \phi}d\phi \right )$$

Unfortunately, I can’t see how I will arrive at the correct expression, ##r^{2}sin\theta drd\theta d\phi ##.

For one reason, when completely expanded, I get terms with repeated differentials like ##dr^{3} ## that don’t cancel.

Why is my method of derivation invalid?

You are not looking at the volumes actually spanned by the basis vectors corresponding to the spherical coordinates. The volume element spanned by three vectors ##\vec a_i## is the triple product ##\vec a_1 \cdot (\vec a_2 \times \vec a_3)##. Hence, if you look at the volume of a small parallelepiped spanned by the coordinate lines with coordinate differences ##dy^i##, then this is given by
$$dV = \frac{\partial \vec x}{\partial y^1} \cdot \left(\frac{\partial \vec x}{\partial y^2} \times \frac{\partial \vec x}{\partial y^3}\right) dy^1 dy^2 dy^3.$$

Alternatively, the volume element can be seen as the wedge product between the differentials. You must then remember that the wedge product is completely anti-symmetric and therefore anything like ##dr\wedge dr = 0##.

quickAndLucky and BvU
Thanks for the help! Using a wedge between the terms in my original expression got me to the answer!

Last edited:

1. What is the formula for deriving the spherical volume element?

The formula for deriving the spherical volume element is: dV = r^2sinθdθdφ, where r is the radius of the sphere, θ is the polar angle, and φ is the azimuthal angle.

2. How is the spherical volume element different from the Cartesian volume element?

The spherical volume element takes into account the curvature of a sphere, while the Cartesian volume element assumes a flat surface. Additionally, the spherical volume element uses polar and azimuthal coordinates, while the Cartesian volume element uses x, y, and z coordinates.

3. What is the purpose of deriving the spherical volume element?

The spherical volume element is used in calculations involving spherical coordinates, such as in physics and engineering problems involving spherical objects or systems. It allows for more accurate and precise calculations in these scenarios.

4. How is the spherical volume element derived?

The spherical volume element is derived using a combination of calculus and geometry. It involves breaking down the volume of a sphere into smaller elements and then integrating them to find the total volume. The final formula is obtained by considering the infinitesimal changes in the polar and azimuthal angles.

5. Are there any real-world applications of the spherical volume element?

Yes, the spherical volume element is used in various fields such as physics, astronomy, and geology. It is used to calculate the volume of planets, stars, and other spherical objects, as well as in fluid dynamics and electromagnetism problems involving spherical symmetry.

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