Problem figuring out the surface of integration

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SUMMARY

The discussion centers on solving a surface integration problem involving a paraboloid intersecting a disk in the xz-plane. The paraboloid intersects the xz-plane at a disk with radius r = √3/3. The user initially attempted integration using polar coordinates with r ranging from 0 to √3/3, but the correct parametrization involves K in polar coordinates, where K is derived from K′ = [0, 3] × [-π/2, π/2]. This clarification resolves the misunderstanding regarding the constraints on y and the correct limits for r.

PREREQUISITES
  • Understanding of surface integration techniques
  • Familiarity with polar coordinates in multivariable calculus
  • Knowledge of paraboloid equations and their geometric properties
  • Basic grasp of integration limits and constraints in calculus
NEXT STEPS
  • Study surface integration methods in multivariable calculus
  • Learn about parametrization techniques for curves and surfaces
  • Explore the properties of paraboloids and their intersections with planes
  • Review polar coordinate transformations and their applications in integration
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Students and professionals in mathematics, particularly those studying calculus, multivariable integration, and geometric analysis. This discussion is beneficial for anyone looking to deepen their understanding of surface integration and parametrization techniques.

Amaelle
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Homework Statement
Consider the vector field F(x, y, z) = (2x, 2y, 2z) and the surface
Σ = n(x, y, z) ∈ R3 : y = 1 − 3x^2 − 3z^2 and x^2 + z^2 ≤ 9, x ≥ 0 ,
oriented so that its normal vector forms an acute angle with the fundamental versor of the y–axis.
Compute the flux of F through Σ.
Relevant Equations
K′ = [0, √3/3] × [-π2, π2 ]
Good day I have a problem figuring out the surface of integration
according to the exercice, we have a paraboloid that cross a disk on the xz plane, the parabloid cross the xz plane on a smaller disk r=√3/3

so for me after going to the final step of integration and using polar coordinate i will integer for a value of r going from 0 to √3/3
but the solution of the exercice said
If we parametrize K in polar coordinates, K is the image of K′ = [0, 3] × [-π2, π2 ]

Any help would be highly appreciated
thanks
 
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Nowhere does it say that y is constrained to be positive; it says instead that x^2 + z^2 \leq 9 which is indeed r \leq 3.
 
  • Informative
Likes   Reactions: Amaelle
Thanks a lot ! I got it now!
 

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