How is this formula for the surface derived?

[musicgold]

Homework Statement

This is not a homework problem. I am struggling with a formula for the inner area of semicircular sphere, mentioned in a book

Relevant Equations

Area = $2 \pi . r . t$

where, r is the radius of the sphere and t is the thickness of its wall.
The formula is shown at the bootom of picture 1.

The pages shown in the pictures are from an engineering book. I am not sure how the thickness of sphere plays a role in the inner surface of the sphere. I know that the surface area of sphere is $4 \pi r^2$ .

Picture 2 shows how that formula plays a role in understanding the stress expereinced by the sphere. I am not sure how the thickness, t, of the sphere wall is relevant here.

Thanks

 

[BvU]

The $2\pi rt$ is the area of the cut surface, according to the text. A ring with radius $r$ and width $t$.

 

[Mark44]

I am not sure how the thickness, t, of the sphere wall is relevant here.

As @BvU already said, they're talking about the area of the cut surface.

Furthermore, the area they give is only approximately correct, and is assuming that the thickness t is small in comparison to the radius r.

The actual area of the cut surface is $\pi r^2 - \pi(r - t)^2 = \pi(r^2 - r^2 - 2rt + t^2) = 2\pi rt - \pi t^2$
If $t << r$, the term $\pi t^2$ won't subtract much from the first term.

 

[musicgold]

Furthermore, the area they give is only approximately correct, and is assuming that the thickness t is small in comparison to the radius r.
The actual area of the cut surface is $\pi r^2 - \pi(r - t)^2 = \pi(r^2 - r^2 - 2rt + t^2) = 2\pi rt - \pi t^2$
If $t << r$, the term $\pi t^2$ won't subtract much from the first term.

Oh, now I get it! Thanks.

Also, I have a a follow up question.

Is it fair to assume that for a sphere with an infinitely thin wall, the inner surface area is equal to the outer surface area?

 

[crashcat]

Is it fair to assume that for a sphere with an infinitely thin wall, the inner surface area is equal to the outer surface area?

Yes.

Although if you're taking the "infinitesimal limit" as in some calculus proof, it won't perfectly cancel and you'll end up with a little differential term which goes to zero in the case that is is literally zero thickness.

 

[kuruman]

The outer and inner surface areas of a shell of outer radius $R$ and inner radius $(R-t)$ are$$A_{outer}=4\pi R^2~;~~~A_{inner}=4\pi (R-t)^2=A_{outer}\left(1-\frac{2t}{R}+\frac{t^2}{R^2}\right).$$ The expression for $A_{inner}$ is exact. You can ignore the quadratic term or both quadratic and linear terms depending on how small the ratio $t/R$ is relative to $1$.