Sphere Calculations: Determining the remaining surface area of the sphere when the sphere is cut by a vertical and horizontal plane

Question

Sphere Calculations: Determining the remaining surface area of the sphere when the sphere is cut by a vertical and horizontal plane and the centre of the sphere is not on the vertical and horizontal axis

I would really appreciate some input regarding a potential solution to this problem I am experiencing with regard to sphere calculations. I am busy creating a model which determines noise levels at various points within a noise sphere. Without going into detail this effectively requires that I need to determine the size of the sphere at every location I took a measurement. While this in itself is not difficult to do even when I include obstructions (like a wall and floor), the problem becomes a lot more difficult when the noise source is not located in the corner of the wall / floor. Consider the first diagram. Here you can see that the noise source is located above the floor at a call “H” and away from the Wall at a length called “L”. The radius of the sphere is known called “R”

The second diagram explains the situation somewhat better.

As can be seen two overlapping spherical caps exist at 90° to one another, to the left and bottom of the two planes shown. Given that I only have the 3 previously mention parameters, how can remaining surface area of the sphere be calculated? I am of course able to determine the surface area of the sphere and each individual spherical cap but am challenged in determining the surface area where the two caps overlap. Any input would be very much appreciated. Before I am asked by a noise expert, it is my intention, by means of numerous noise measurements, to determine the amount of absorption or reflection of the wall and floor.

A similar challenge will also be forthcoming when a second wall is added resulting in larger than normal 8th sphere with the added dimension of L2.

Answer 1

Sphere: $\rho=R$

Horizontal plane: $x=\rho\cos\theta\sin\varphi=x_0$

Vertical plane: $z=\rho\cos\varphi=z_0$

Area:

$$ A=\int_0^{\arccos(R/z_0)} \int_{\arcsin(x_0\csc\varphi/R)}^{2\pi-\arcsin(x_0\csc\varphi/R)} R^2\sin\varphi\,d\theta d\varphi. $$

Answer 2

The integral mentioned above dated 2014-03-19 seems to be completely solvable in terms of rational functions and inverse trigonometric functions: $$ A = R^2\int_0^{\arccos(R/z_0)} \sin\varphi d\varphi \int_{\arcsin (x_0\csc \varphi/R)}^{2\pi - \arcsin(x_0\csc \varphi/R)}d\theta $$ $$ = R^2\int_0^{\arccos(R/z_0)} \sin\varphi d\varphi [2\pi-2\arcsin (x_0\csc \varphi/R)] $$ $$ = 2\pi R^2\int_0^{\arccos(R/z_0)} \sin\varphi d\varphi -2R^2\int_0^{\arccos(R/z_0)} \sin\varphi d\varphi \arcsin (x_0\csc \varphi/R) $$ $$ = 2\pi R^2(1-R/z_0) -2R^2\int_0^{\arccos(R/z_0)} \sin\varphi d\varphi \arcsin [ (x_0/R)/\sin \varphi] $$ $$ = 2\pi R^2(1-R/z_0) -2R^2[\bar A(\arccos (R/z_0))-\bar A(0)] . $$ In the second term, the substitution $\frac{x_0}{R\sin \varphi}=u$ gives $$ \sin\varphi = \frac{x_0}{Ru},$$ $$ du/d\varphi = -\frac{x_0\cos\varphi}{R \sin^2\varphi},$$ $$ \sin^2\varphi du = -\frac{x_0\cos\varphi}{R}d\varphi,$$ $$ \sin^3\varphi du = -\frac{x_0\cos\varphi}{R}\sin\varphi d\varphi,$$ $$ \bar A(\varphi)=\int \sin\varphi d\varphi \arcsin ( (x_0/R)/\sin \varphi) $$ $$ = -\int \frac{R}{x_0}\frac{\sin^3\varphi}{\cos \varphi} \arcsin u du $$ $$ = -\int \frac{x_0^2}{R^2}\frac{1}{u^3}\frac{1}{\cos \varphi} \arcsin u du $$ $$ = -\frac{x_0^2}{R^2}\int \frac{1}{u^3\sqrt{1-\frac{x_0^2}{R^2u^2}}} \arcsin u du $$ $$ = -\frac{x_0^2}{R^2}\int \frac{1}{u^2\sqrt{u^2-\frac{x_0^2}{R^2}}} \arcsin u du = -\frac{x_0^2}{R^2}I_{x_0/R}(u) . $$ This integral is solved by partial integration with $\frac{d}{du}\arcsin u= \frac{1}{\sqrt{1-u^2}}$ and $\int \frac{1}{u^2\sqrt{u^2-a^2}}du=\frac{\sqrt{u^2-a^2}}{a^2u}$: $$ I_a(u)=\int \frac{1}{u^2\sqrt{u^2-a^2}} \arcsin u du = \frac{\sqrt{u^2-a^2}}{a^2u}\arcsin u - \int \frac{1}{\sqrt{1-u^2}} \frac{\sqrt{u^2-a^2}}{a^2u} du. $$ The substitution with $u^2=v$, $dv=2udu$ yields $$ J_a(u)=\int \frac{1}{\sqrt{1-u^2}} \frac{\sqrt{u^2-a^2}}{a^2u} du = \frac{1}{2a^2} \int \frac{1}{\sqrt{1-v}} \frac{\sqrt{v-a^2}}{v} dv = \frac{1}{2a^2} \int \frac{v-a^2}{\sqrt{1-v}} \frac{1}{\sqrt{v-a^2}v} dv $$ $$ = \frac{1}{2a^2} \int \frac{1}{\sqrt{1-v}} \frac{1}{\sqrt{v-a^2}} dv - \frac{1}{2} \int \frac{1}{\sqrt{1-v}} \frac{1}{v\sqrt{v-a^2}} dv. $$ With the Gradsteyn-Rzyhik table of integrals formula 2.261 $$ \int\frac{1}{\sqrt{1-v}\sqrt{v-a^2}}dv = -\arcsin\frac{-2v+1+a^2}{1-a^2} $$ and with the Gradsteyn-Ryzhik formula 2.266 $$ \int\frac{1}{v\sqrt{1-v}\sqrt{v-a^2}}dv =\frac{1}{a} \arcsin\frac{-2a^2+(1+a^2)v}{v(1-a^2)}. $$ In conclusion $J_a$ is an expression of 2 arcsines, $I_a$ and $\bar A$ are expressions of 3 arcsines.