Can You Recall the Formula for Finding the Area of a Spherical Patch?

[Curl]

Does anyone remember the formula for the area of a spherical patch in terms of two angles?

Obviously you parametrize the surface and do the surface integral but I'm a bit too lazy/busy right now. So does anyone just remember the result?

By spherical patch I mean something like this:

Spoiler

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I want to define it by two angles which are between some arbitrary values. The z-axis passes through the center of the patch and is normal to the surface at the intersection point.

 

[HallsofIvy]

The "differential of surface area" of a sphere of radius R is $R^2 sin(\phi)d\phi d\theta$ so the area of a spherical patch with $\theta$ between $\theta_0$ and $\theta_1$, $\phi$ between $\phi_0$ and $\phi_1$ is given by
$$\int_{\theta= \theta_0}^{\theta_1}\int_{\phi= \phi_0}^{\phi_1} R^2 sin(\phi)d\phi d\theta$$
$$= R^2(\theta_1- \theta_0)(cos(\phi_0)- cos(\phi_1))$$

 

[dharasty]

I have the impression that HallsofIvy did not answer Curl's question... however, I'm extrapolating a bit based on Curl's verbiage and picture.

I think Curl is depicting what would be the intersection of two orthogonal lunes, that is: all for "sides" of the patch are segments along great circles. (I'd informally call this a "rectangular patch of a sphere"...)

I think HallsofIvy answered as if the "sides" of the patch are like "latitude and longitude lines" on the globe. As we know, latitude lines are NOT great circles, except for latitude = 0°. (I'd informally call this a "trapezoidal patch of a sphere"...)

So I think Curl's question is yet to be answered.

However -- if I'm right -- then I got the answer I was looking for, which was the "latitude-longitude" sense.

 

[awkward]

If the sides of the "patch" are segments of great circles, then the area on the surface of a sphere of radius 1 is given by the "pyramid solid angle formula",
$$\Omega = 4 \arcsin \left( \sin \frac{a}{2} \sin \frac{b}{2} \right)$$
where a and b are the apex angles. Mutiply by $R^2$ for a sphere of radius $R$.

See the section titled "Pyramid" in
http://en.wikipedia.org/wiki/Solid_angle

 

[CellCoree]

The formula for the area of a spherical patch in terms of two angles is given by:

A = R^2 * (θ2 - θ1) * (sin φ2 - sin φ1)

Where R is the radius of the sphere, θ1 and θ2 are the two angles defining the boundaries of the patch, and φ1 and φ2 are the corresponding angles on the vertical axis.

This formula can be derived by parametrizing the surface of the spherical patch and performing a surface integral, as mentioned. However, it is important to note that the result may vary depending on the orientation and parametrization chosen.

I hope this helps. If you need further assistance, please let me know.