How Do You Calculate the Intersection of Two Spheres in 3D Integrals?

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SUMMARY

The discussion focuses on calculating the intersection of two spheres in three-dimensional integrals, specifically using the inequalities C=|\vec{s}-\vec{t}|<1 and |\vec{s}+\vec{t}|<1. The user seeks strategies for solving these inequalities for fixed \vec{s} using spherical coordinates. A suggested approach is to utilize cylindrical-polar coordinates with the z-axis perpendicular to the plane of intersection, treating the problem as a volume of rotation.

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  • Knowledge of vector operations in 3D
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kassem84
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Hello,
I am calculating some integrals in 3 dimensions. However, the difficulties of such integrals lie in the determination of the boundaries of the variables integrated over.

[itex]\int_{C} d^{3}\vec{t}[/itex] e[itex]^{-\vec{s}.\vec{t}}[/itex]
For example, if we consider (C) as the region of the intersection of 2 spheres:
C=|[itex]\vec{s}[/itex]-[itex]\vec{t}[/itex]|<1 and |[itex]\vec{s}[/itex]+[itex]\vec{t}[/itex]|<1
I want to solve these set of inequalities for fixed [itex]\vec{s}[/itex], using spherical coordinates.
i.e. determine the interval over |[itex]\vec{t}[/itex]|, [itex]\phi[/itex] and [itex]\vartheta[/itex]=angle([itex]\vec{s}[/itex],[itex]\vec{t}[/itex])

Does anyone have a strategy to deal with such inequalities?

Thanks in advance.[itex]^{}[/itex]
 
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Use cylindrical-polar coordinates with the z axis perpendicular to the plane of intersection.
You can treat it as a volume of rotation.
 

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