Finding the Volume of a tetrahedron using Spherical Coordinates

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SUMMARY

The volume of a tetrahedron under the plane defined by the equation 3x + 2y + z = 6 in the first octant can be calculated using spherical coordinates. The correct volume is determined to be six. The spherical coordinate transformations are given by x = p sin(φ) cos(θ), y = p sin(φ) sin(θ), and z = p cos(φ). The boundaries for the spherical coordinates are established as p ranging from 0 to (6/(cos(φ) + sin(φ)(3cos(θ) - 2sin(θ))), with θ between 0 and π/2 and φ between 0 and π/2.

PREREQUISITES
  • Understanding of spherical coordinates
  • Familiarity with the equation of a plane
  • Knowledge of triple integrals for volume calculation
  • Basic calculus concepts
NEXT STEPS
  • Study the derivation of volume formulas using spherical coordinates
  • Learn about the application of triple integrals in different coordinate systems
  • Explore the concept of boundaries in integration for geometric shapes
  • Review the transformation equations between Cartesian and spherical coordinates
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on calculus, geometry, and applications of spherical coordinates in volume calculations.

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Find the volume of a tetrahedron under a plane with equation 3x + 2y + z = 6 and in the first octant. Use spherical coordinates only. The answer is six.




x=psin(phi)cos(theta)
y=psin(phi)sin(theta)
z=pcos(phi)



I've been trying to figure out the boundaries of this particular problem all night. To be honest, I'm completely at a loss. I have p going between 0 and (6/(cos(phi)+sin(phi)(3cos(theta)-2sin(theta)). I believe that theta is between 0 and pi/2, although I'm not entierly sure on that one. As far as phi goes I believe the upper limit is pi but the lower limit is a mystery to me.
 
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\phi would go from 0 to \pi/2 as does \theta.
 

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