# Area of a spherical cap across an intersection [duplicate]

I have a sphere with radius $$r$$ that equally divided in to two hemisphere P and S. There is a plane that separate two different zone A and B. Upper zone called A and lower zone called B. $$\alpha$$ is the contact angle. So $$\alpha$$ larger means sphere is more in zone A and vice versa. Range of $$\alpha$$ can go between $$[0,\pi]$$The angle $$\beta$$ defines the orientation of the sphere which is an angle between z axis (vertical) and a unit vector. Range of $$\beta$$ can be $$[0,2\pi]$$. The Schematic representation of this is below.

To calculate the area of $$P$$ portion of the sphere into $$A$$ is given by following equation $$\mathrm{Area}_{P@A}= r^2\int_{\theta=\frac{\pi}{2}-\beta}^\alpha \int_{\phi=\arcsin(1/(\tan\theta \tan \beta))} ^{\pi -\arcsin(1/(\tan\theta \tan \beta)} \sin\theta\; d\theta d\phi.$$

After solving this equation we can get the following solution. $$\textrm{Area}_{P@A}= 2 r^2 \left\{ \cos (\alpha) \sin ^{-1}(\cot (\alpha) \cot (\beta)) -\tan ^{-1}\left(\frac{\cos (\beta)}{\sqrt{\sin ^2(\beta)-\cos ^2(\alpha)}}\right)\right\}+\pi r^2 (1-\cos (\alpha)).$$

Note that the solution is defined as a function of $$\alpha$$ and $$\beta$$. If we change the value of $$\alpha$$ and $$\beta$$, the area of P into A will change eventually.

Now consider more complex case where $$P$$ and $$S$$ are not equal hemisphere.

Question: If $$P$$ and $$S$$ are not equal hemisphere rather they are spherical cap, then what I need to consider to calculate the area of $$P$$ into zone $$A$$ as a function of $$\alpha$$ and $$\beta$$? Here $$\gamma$$ defines size of the spherical cap and r1 is the radius of the base of the cap. Do I need to introduce any other parameter?

It is confusing that you use $$\theta$$ for both the azimuthal coordinate, and the offset of the spherical cap. Let us change the notation, so the offset is described by an angle $$\gamma$$ in your second figure. You should also label your axis, let's say the vertical axis is $$z$$, and the horizontal one is $$x$$, while $$y$$ is perpendicular to the page. With these definitions, the equation for the boundary of the spherical cap is $$z = x\tan \beta - r\sin \gamma$$ while the points on the sphere are described by $$z = r\cos \theta$$ $$x = r\sin \theta \sin \phi$$ $$y = r\sin \theta \cos \phi$$

Plugging one in the other, we find the relationship between $$\theta$$ and $$\phi$$: $$\cos \theta = \sin \theta \sin \phi \tan \beta - \sin \gamma$$

The smallest $$\theta$$, or the tip of the cap, is found when $$\phi = \pi/2$$: $$\cos \theta_0 = \sin \theta_0 \tan \beta - \sin \gamma$$ from which we find $$\theta_0 = -\beta + \cos^{-1} (-\cos \beta \sin \gamma)$$

So the area is $$A/r^2 = \int_{-\beta + \cos^{-1} (-\cos \beta \sin \gamma)}^{\alpha} d\theta\; \sin \theta \left(\pi - 2\sin^{-1} \left[\frac{\cos \theta+\sin \gamma}{\sin \theta \tan \beta} \right] \right)$$