Intersection of spherical shell and a plane

[mae3x]

I need to know the area of the intersection between a sperical shell and a plane in spherical coordinates. By "shell" I mean a sphere with some differential thickness dR. Basically, I know that the intersection of a sphere and a plane is a circle. But I want to consider this sphere having some differential thickness and determine the subsequent differential area of intersection with the plane. Can someone please help me out with formulating this problem? Thanks, Mike

 

[quasar987]

Isn't this just a ring with differential tickness dR?

Consider a ring of inner radius r and outer rarius R.
$$A_{ring} = \pi(R^2-r^2)$$

How does its area varies with an increase of R?

$$\frac{dA_{ring}}{dR} = \pi (2R-r^2)$$

So a differential increase of dR means a differential increase in the area of

$$dA_{ring} = \pi (2R-r^2)dR$$

And if the outer radius is r (R=r), then you have the concept of a diffential ring of radius r, i.e. what you're looking for:

$$dA_{ring}(R=r) = \pi (2r-r^2)dR$$

 

[D H]

$$dA_{ring}(R=r) = \pi (2r-r^2)dR$$

That solution is incorrect upon inspection as it involves the difference between an length and an area. The correct answer is $2\pi r dR$.

The error appeared by taking [itex]\frac{d}{dR} \pi (R^2-r^2)= \pi (2R-r^2)[/tex], which is incorrect: $\frac{dr}{dR} = 0$.

 

[HallsofIvy]

Isn't this just a ring with differential tickness dR?

Consider a ring of inner radius r and outer rarius R.
$$A_{ring} = \pi(R^2-r^2)$$

How does its area varies with an increase of R?

$$\frac{dA_{ring}}{dR} = \pi (2R-r^2)$$

If you are assuming that r is a constant then
$$\frac{dA_{ring}}{dR} = 2\pi R$$
since the derivative of r² is 0.

So a differential increase of dR means a differential increase in the area of

$$dA_{ring} = \pi (2R-r^2)dR$$

And if the outer radius is r (R=r), then you have the concept of a diffential ring of radius r, i.e. what you're looking for:

$$dA_{ring}(R=r) = \pi (2r-r^2)dR$$

 

[quasar987]

oops.