How Does Deformation Affect Surface and Volume Elements on a Sphere?

[gasar8]

Homework Statement

Show that the surface and volume element on a deformed sphere are
$$\sigma = \frac{\rho^2 \sin \theta}{\cos \gamma} d\phi d\theta,$$
$$dV = \rho^3 \sin \theta d\phi d\theta,$$
if $\gamma$ is the angle between normal vector and radius vector.

Homework Equations

$$n\cdot r = \cos \gamma,$$
$$dS = \bigg| \frac{\partial X}{\partial \phi} \times \frac{\partial X}{\partial \theta}\bigg| ,$$
where $X = (r \sin \theta \cos \phi, r \sin \theta \sin \phi, r \cos \theta)$.

The Attempt at a Solution

I assume that this cross product above is normal vector by definition, so I think it must hold that
$$n = (-r^2 \cos\phi \sin^2\theta, -r^2 \sin^2 \theta \sin \phi, -r^2 \cos \theta \sin \theta),$$
so its norm would be $r^2 \sin \theta$, which is similar to what my task is, but I do not know where to put $\cos \gamma$ so that it would be really equal.
It seems unlogical to me, if $\vec{r} || \vec {n}$ then $\cos \gamma =1$, but if they are perpendicular the surface element goes to infinity?

VOLUME ELEMENT:
From the sketch that I draw, I would again assume that the triple product
$$(\frac{\partial X}{\partial \phi} \times \frac{\partial X}{\partial \theta}) \cdot X$$
would give me the volume of the parallelepiped spanned between $d \phi, d\theta \ \text{and} \ r,$ and since I am looking for some kind of pyramid with tilted base surface, I assume that the volume is than the third of this parallelepiped, but in the result I am searching for, there is no 3?

EDIT: On the other hand if I draw everything up, it seems logical that the projected plane is just $\cos \gamma$ times the original surface area, which is $r^2 \sin \theta$, but is there a way to derive it using Jaccobian?

 

[mark726]

Thank you for your forum post. I understand that you are trying to show that the surface and volume elements on a deformed sphere are given by the equations \sigma = \frac{\rho^2 \sin \theta}{\cos \gamma} d\phi d\theta and dV = \rho^3 \sin \theta d\phi d\theta, respectively. I will attempt to guide you through the solution and provide some clarification on your attempted solution.

Firstly, I would like to address your attempt at finding the normal vector \vec{n}. The normal vector is defined as the vector perpendicular to the surface at a given point. In this case, the surface is the deformed sphere and the point is specified by the coordinates (r, \phi, \theta). The normal vector can be found by taking the cross product of the two tangent vectors to the surface, \frac{\partial X}{\partial \phi} and \frac{\partial X}{\partial \theta}. This gives us the normal vector \vec{n} = \frac{\partial X}{\partial \phi} \times \frac{\partial X}{\partial \theta} = (-r \sin \theta \sin \phi, r \sin \theta \cos \phi, 0).

Now, let's consider the angle \gamma between the normal vector and the radius vector \vec{r} = (r \sin \theta \cos \phi, r \sin \theta \sin \phi, r \cos \theta). As you correctly pointed out, if the normal vector and the radius vector are perpendicular, then \cos \gamma = 0 and the surface element goes to infinity. However, this is not the case for a deformed sphere. The normal vector and the radius vector are not necessarily perpendicular, but they are related by the dot product \vec{n} \cdot \vec{r} = \cos \gamma. This means that the angle \gamma varies with the position on the surface, and it is not a constant value.

Next, let's look at the surface element dS. This is given by the magnitude of the cross product of the two tangent vectors, which we have already found to be \vec{n}. Therefore, we have dS = |\vec{n}| = r^2 \sin \theta. This is the same as the surface area of a unit sphere, which is expected since we