Triple integral volume of solid (set up)

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SUMMARY

The discussion focuses on setting up a triple integral in spherical coordinates to calculate the volume of a solid defined by the region above the xy-plane, inside the sphere defined by the equation x² + y² + z² = 2, and outside the cylinder x² + y² = 1. The correct integral setup involves two integrals: the first for the volume inside the sphere and the second for the volume of the cylinder, both expressed in spherical coordinates. The user initially misidentified the cylinder's equation but later corrected it to x² + y² = 1, confirming the need for proper limits in the integration process.

PREREQUISITES
  • Understanding of spherical coordinates and their application in triple integrals.
  • Knowledge of the equations of spheres and cylinders in three-dimensional space.
  • Familiarity with the concept of volume integration in calculus.
  • Ability to manipulate and set up multiple integrals correctly.
NEXT STEPS
  • Study the derivation and application of spherical coordinates in triple integrals.
  • Learn how to convert Cartesian equations of solids into spherical coordinates.
  • Practice setting up and evaluating triple integrals for various solid regions.
  • Explore the use of Jacobians in changing variables for multiple integrals.
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus and integration techniques, as well as anyone needing to understand volume calculations in three-dimensional geometry.

whynot314
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Homework Statement


I need help setting this integral up in spherical coordinates, the region above the xyplane, inside the sphere x^2+y^2+z^2=2 and outside the cylinder x^2+y^2=2





The Attempt at a Solution




\int^{2\pi}_{0} \int^{\pi/2}_{\pi/4} \int^{\sqrt{2}}_{0} \rho^{2}sin\phid\rhod\thetad\theta -\int^{2\pi}_{0} \int^{\pi/4}_{0} \int^{1/sin\phi}_{0} \rho^{2}sin\phid\rhod\phid\theta

is this correct?
 
Physics news on Phys.org
cylinder is x^2+y^2=1 oops, and on first integrand its drho,dphi,dtheta.
 

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