Steradian and Radian relation, Surface Area and Volume of Sphere, Projected Area of region on curved surface of a sphere

To find the relation between 3D Steradian and 2D Radian of a Spherical cap region

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Small Elemental area dA is on the curved surface of the sphere, \[\text{C.S.A. of spherical cap}= \int_0^\theta dA\] \[= \int 2\pi r\,dr\] \[= \int 2\pi(R\sin \theta)(R\,d\theta)\] \[= 2\pi R^2\int_0^\theta \sin \theta \, d\theta\] \[= 2\pi R^2(1-\cos\theta)\] We know that, \[\text{C.S.A. of spherical cap}=\Omega \,R^2\]

\[\boxed{\Omega = 2\pi(1-\cos\theta)}\] Note that the $\theta$ is the half angle in 2D Radians that corresponds to the 3D $\Omega$

Projected Area for Dot Product

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Projected area of the small elemental area dA is $ dA cos \theta$, For a region from 0 to $\theta $ \[\text{Projected Area of Spherical Cap}= \int_0^\theta dA\] \[= \int 2\pi r\, dr \,\cos\theta\] \[= \int 2\pi(R\sin \theta)(R\,d\theta)\,\cos\theta\] \[= \pi R^2\int_0^\theta \sin 2\theta \, d\theta\] \[= \pi R^2\left[\frac{-\cos 2\theta}{2}\right]_0^\theta \]

\[\boxed{Projected \,Area = \frac{\pi R^2}{2}\left(1-\cos2\theta\right)}\]

Same procedure can be followed if projected area is required between specific angles such as $\theta_1$ to $\theta_2$ Also note that: Projected area is in vertical plane perpendicular to the paper. But the area vector for it will be along positive X direction which agrees with $dA \,cos\,\theta$ direction. $\\\\$ Method 2: \[\text{Projected Area of Spherical Cap}= \pi r^2\] \[= \pi \left(R\sin\theta\right)^2\] \[Projected \,Area = \frac{\pi R^2}{2}\left(1-\cos2\theta\right)\]

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Surface Area of Sphere

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Small Elemental area dA is on the curved surface of the sphere, \[\text{C.S.A. of spherical cap}= \int_0^\pi dA\] \[= \int 2\pi r\,dr\] \[= \int 2\pi(R\sin \theta)(R\,d\theta)\] \[= 2\pi R^2\int_0^\pi\sin \theta \, d\theta\] \[= 2\pi R^2\left[-\cos\theta\right]_0^\pi\]

\[\boxed{Surface\, Area \,of \,Sphere = 4\pi R^2}\]

Volume of Sphere

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Small Elemental Volume dV in the shaded region inside the sphere, \[dV = \int 4\pi r^2\,dr\] \[V=\int_0^R dV \] \[ = 4\pi \int_0^R r^2 dr\] \[ =4\pi \left[ \frac{r^3}{3}\right]_0^R\]

\[\boxed{ Volume \,of \,Sphere = \frac{4}{3}\pi R^3 }\]