the surface area of the cream white colored surface wants to be calculated using integral

Question

I Want to calculate the area of the cream colored surface illustrated on the image below using integral. variables are $\beta$ and $\phi$ and constants are R and r

Answer 1

Some thoughts.

I use spherical coordinates \begin{align} x&=R\sin\theta\cos\varphi\\ y&=R\sin\theta\sin\varphi\\ z&=R\cos\theta \end{align} where my $\theta$ corresponds to $\varphi$ in your image.
The unit versor normal to the plane inclined at an angle $\beta$ is $$ \mathbf{n}=\cos\beta\,\mathbf{j}+\sin\beta\,\mathbf{k} $$ so that the equation of the inclined plane (passing through $(0,0,h)$) is $$ y\cos\beta+(z-h)\sin\beta=0 $$ Here $h$ is $R-r=R\cos\theta_0$, and $\theta_0$ is the maximum value of $\theta$ corresponding to the horizontal plane $z=R-r$.

The equation of the inclined plane in spherical coordinates becomes $$ R\sin\theta\sin\varphi\cos\beta+(R\cos\theta-h)\sin\beta=0 $$ This is a contraint between $\theta$, $\varphi$, we can write $$ \sin\varphi=-\frac{\cos\theta-\cos\theta_0}{\sin\theta}\tan\beta $$ This sets a minimum value for $\sin\varphi$, except for the higher part of the surface, where the RHS of last eq. is lesser than $-1$, say for $0\leq\theta\leq\theta_1$. Set, where meaningful, $$ \varphi(\theta)=\arcsin\left(\frac{\cos\theta-\cos\theta_0}{\sin\theta}\tan\beta\right) $$ so that we can write ($R^2\sin\theta\,d\theta\,d\varphi$ is the surface element in spherical coordinates): \begin{align} S&=\int_Sd\sigma=\int_0^{\theta_1}d\theta\int_0^{2\pi}R^2\sin\theta\,d\varphi +\int_{\theta_1}^{\theta_0}d\theta \int_{-\varphi(\theta)}^{\pi+\varphi(\theta)}R^2\sin\theta\,d\varphi\\ &=2\pi R^2(1-\cos\theta_1) +R^2\int_{\theta_1}^{\theta_0}(\pi+2\varphi(\theta))\sin\theta\,d\theta \end{align} I cannot go further in calculating the last integral.