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

In summary: Finally, let's consider the volume element dV. This can be found by taking the triple product of the three tangent vectors, \frac{\partial X}{\partial \phi}, \frac{\partial X}{\partial \theta}, and \vec{r}. This is given by dV = \bigg| \frac{\partial X}{\partial \phi} \times \frac{\partial X}{\partial \theta} \cdot \vec{r} \bigg| = r^3 \sin \theta. This is the same as the volume of a unit sphere, which again aligns with our expectations for a deformed sphere.In summary, we have shown that the
  • #1
gasar8
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Homework Statement


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

Homework Equations


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

The Attempt at a Solution


I assume that this cross product above is normal vector by definition, so I think it must hold that
[tex] n = (-r^2 \cos\phi \sin^2\theta, -r^2 \sin^2 \theta \sin \phi, -r^2 \cos \theta \sin \theta),
[/tex]
so its norm would be [itex] r^2 \sin \theta [/itex], which is similar to what my task is, but I do not know where to put [itex] \cos \gamma [/itex] so that it would be really equal.
It seems unlogical to me, if [itex] \vec{r} || \vec {n}[/itex] then [itex] \cos \gamma =1 [/itex], 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
[tex] (\frac{\partial X}{\partial \phi} \times \frac{\partial X}{\partial \theta}) \cdot X
[/tex]
would give me the volume of the parallelepiped spanned between [itex] d \phi, d\theta \ \text{and} \ r, [/itex] 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 [itex]\cos \gamma [/itex] times the original surface area, which is [itex] r^2 \sin \theta [/itex], but is there a way to derive it using Jaccobian?
 
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  • #2

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
 

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

1. What is a surface element?

A surface element is a small portion of a larger surface that is used to calculate the total surface area. It is typically represented as dS and is a vector quantity with direction perpendicular to the surface.

2. How is the surface element calculated?

The surface element is calculated by taking the cross product of two vectors tangent to the surface at a specific point. This gives the direction and magnitude of the surface element.

3. What is the significance of the surface element in calculus?

In calculus, the surface element is used to calculate the surface integral, which is a way of measuring the flux or flow through a surface. It is also used in the calculation of surface area and surface integrals in vector calculus.

4. What is a volume element?

A volume element is a small portion of a larger volume that is used to calculate the total volume. It is typically represented as dV and is a scalar quantity with magnitude equal to the volume of the element.

5. How is the volume element used in multivariable calculus?

In multivariable calculus, the volume element is used in the calculation of triple integrals, which are used to find the volume of a three-dimensional object. It is also used in the calculation of surface integrals in vector calculus.

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