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Sphere-Sphere Intersection

\begin{figure}\BoxedEPSF{SphereSphereIntersection.epsf scaled 700}\end{figure}

Let two spheres of Radii $R$ and $r$ be located along the x-Axis centered at $(0,0,0)$ and $(d,0,0)$, respectively. Not surprisingly, the analysis is very similar to the case of the Circle-Circle Intersection. The equations of the two Spheres are

$\displaystyle x^2+y^2+z^2$ $\textstyle =$ $\displaystyle R^2$ (1)
$\displaystyle (x-d)^2+y^2+z^2$ $\textstyle =$ $\displaystyle r^2.$ (2)

Combining (1) and (2) gives
\end{displaymath} (3)

Multiplying through and rearranging give
\end{displaymath} (4)

Solving for $x$ gives
x={d^2-r^2+R^2\over 2d}.
\end{displaymath} (5)

The intersection of the Spheres is therefore a curve lying in a Plane parallel to the $yz$-plane at a single $x$-coordinate. Plugging this back into (1) gives
$\displaystyle y^2+z^2$ $\textstyle =$ $\displaystyle R^2-x^2=R^2-\left({d^2-r^2+R^2\over 2d}\right)^2$  
  $\textstyle =$ $\displaystyle {4d^2R^2-(d^2-r^2+R^2)^2\over 4d^2},$ (6)

which is a Circle with Radius

$\displaystyle a$ $\textstyle =$ $\displaystyle {1\over 2d}\sqrt{4d^2R^2-(d^2-r^2+R^2)^2}$  
  $\textstyle =$ $\displaystyle {1\over 2d}[(-d + r - R)(-d - r + R)(-d + r + R)(d + r + R)]^{1/2}.$ (7)

The Volume of the 3-D Lens common to the two spheres can be found by adding the two Spherical Caps. The distances from the Spheres' centers to the bases of the caps are

$\displaystyle d_1$ $\textstyle =$ $\displaystyle x$ (8)
$\displaystyle d_2$ $\textstyle =$ $\displaystyle d-x,$ (9)

so the heights of the caps are
$\displaystyle h_1$ $\textstyle =$ $\displaystyle R-d_1={(r-R+d)(r+R-d)\over 2d}$ (10)
$\displaystyle h_2$ $\textstyle =$ $\displaystyle r-d_2={(R-r+d)(R+r-d)\over 2d}.$ (11)

The Volume of a Spherical Cap of height $h'$ for a Sphere of Radius $R'$ is
V(R',h')={\textstyle{1\over 3}}\pi h'^2(3R'-h').
\end{displaymath} (12)

Letting $R_1=R$ and $R_2=r$ and summing the two caps gives

V=V(R_1, h_1)+V(R_2,h_2) = {\pi(R+r-d)^2(d^2+2dr-3r^2+2dR+6rR-3R^2)\over 12d}.
\end{displaymath} (13)

This expression gives $V=0$ for $d=r+R$ as it must. In the special case $r=R$, the Volume simplifies to
V={\textstyle{1\over 12}}\pi(4R+d)(2R-d)^2.
\end{displaymath} (14)

See also Apple, Circle-Circle Intersection, Double Bubble, Lens, Sphere

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© 1996-9 Eric W. Weisstein