Volume using double integral (polar coordiantes)

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SUMMARY

The discussion focuses on calculating the volume of a solid bounded by the surfaces defined by the equations z = x² + 2y² and z = 12 - 2x² - y² using double integrals in polar coordinates. The intersection of these surfaces occurs at a circle with a radius of 2. The correct setup for the double integral is confirmed as ∫₀²π ∫₀² (12 - 3r²)r dr dθ, which accurately represents the volume calculation by integrating the difference between the upper and lower surfaces.

PREREQUISITES
  • Understanding of double integrals
  • Knowledge of polar coordinates
  • Familiarity with volume calculations in multivariable calculus
  • Ability to manipulate and interpret equations of surfaces
NEXT STEPS
  • Study the application of double integrals in cylindrical coordinates
  • Learn how to derive the equations of surfaces from given conditions
  • Explore advanced techniques for volume calculations using triple integrals
  • Investigate the geometric interpretation of double integrals in three-dimensional space
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus and volume calculations, as well as anyone looking to deepen their understanding of polar coordinates in integration.

whynot314
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Homework Statement


use a double integral to find the volume of the solid bounded by.

z=x^2+2y^2 and z=12-2x^2-y^2

I want to change variables using polar coordinates, I know its the top minus the bottom, and the intersection between the two is a circle radius 2.

The Attempt at a Solution



I want to make sure i have the correct set up

[itex]\int^{2\pi}_{0}[/itex] [itex]\int^{2}_{0}[/itex] (12-3r[itex]^{2}[/itex])rdrd[itex]\theta[/itex]
 
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whynot314 said:

Homework Statement


use a double integral to find the volume of the solid bounded by.

z=x^2+2y^2 and z=12-2x^2-y^2

I want to change variables using polar coordinates, I know its the top minus the bottom, and the intersection between the two is a circle radius 2.

The Attempt at a Solution



I want to make sure i have the correct set up

[itex]\int^{2\pi}_{0}[/itex] [itex]\int^{2}_{0}[/itex] (12-3r[itex]^{2}[/itex])rdrd[itex]\theta[/itex]
That looks good !
 

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