Triple integral w/ spherical subsitution

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SUMMARY

The discussion focuses on computing the derivative F'(t) of the function F(t) defined by the triple integral of a differentiable function f(x) over a spherical region. The spherical coordinates used are x = p sin φ cos θ, y = p sin φ sin θ, and z = p cos φ, with the bounds 0 < p < t, 0 < φ < π, and 0 < θ < 2π. The Jacobian determinant for the transformation is p² sin φ. The integral simplifies due to the independence of f(p²) from φ, allowing for straightforward evaluation of the φ integral.

PREREQUISITES
  • Understanding of triple integrals in spherical coordinates
  • Knowledge of Jacobian determinants for coordinate transformations
  • Familiarity with differentiable functions and their properties
  • Basic integration techniques, particularly involving trigonometric functions
NEXT STEPS
  • Study the evaluation of triple integrals in spherical coordinates
  • Learn about Jacobian determinants and their applications in multivariable calculus
  • Explore the properties of differentiable functions in the context of integration
  • Practice integrating trigonometric functions, specifically sin(φ) over defined intervals
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Students and educators in calculus, particularly those focusing on multivariable calculus and integration techniques, as well as anyone seeking to deepen their understanding of spherical coordinate transformations.

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Homework Statement


f(x) is a differentiable function let
F(t)= \int\int\int_{x^2+y^2+z^2\leq t^2} f(x^2+y^2+z^2) dx dy dz

compute F^{&#039;}(t)

Homework Equations



x=p sin \phi cos\theta
y= p sin \phi sin\theta
z= p cos \phi

spherical bounds 0<p<t 0<\phi<\Pi 0<\theta < 2\Pi

p^2 sin\phi = jacobian determinant

3. The attempt at a solution

carried through the substitution \int\int\int f(p^2) p^2 sin \phi dp d\phi d\theta

dont know how to evaluate \intf(p^2) sin\phi d\phi?




 
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Be careful about putting those repeated copies of the template in your questions.

The remaining integral is trivial. Your function is independent of \phi so the \phi integration is trivial; it's just the integral of sin(\phi)
 

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