Help with triple integration problem

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Discussion Overview

The discussion revolves around a triple integration problem using spherical coordinates, specifically calculating the volume of a small part of a sphere defined by certain limits for the variables ρ, φ, and θ. The focus is on verifying the correctness of an initial answer and clarifying the integration process involved.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents an initial answer for the volume calculation using spherical coordinates and seeks confirmation of its correctness.
  • Another participant proposes a formula for the volume based on the given limits and suggests that the initial answer could be a simplification of this formula.
  • A later reply questions the order of cosine terms in the integration limits, prompting a clarification on the integration process.
  • A subsequent response confirms the reasoning behind the cosine terms and acknowledges a mistake in the initial understanding.

Areas of Agreement / Disagreement

Contextual Notes

Participants rely on specific integration techniques and properties of trigonometric functions, which may depend on their interpretations of the limits and the integration process.

mister_okay
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hey! i need some help with a triple integration problem using spherical coordinates. it's the volume of a small part of a sphere. rho from 5 to 6, phi from pi/6 to pi/4 and theta from pi/4 to pi/3.

i got an answer of (-91/72) x pi x (sqrt(2)-sqrt(3))...am i right? Thanks!
 
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Well, the answer ought to be:
[tex]\frac{6^{3}-5^{3}}{3}(\frac{\pi}{3}-\frac{\pi}{4})(\cos\frac{\pi}{6}-\cos\frac{\pi}{4})[/tex]
So, if your answer is just a simplification of this, then it is correct.


EDIT:
Yup, it is correct.
 
Last edited:
Hey thanks arildno :)

but one question..

shouldnt it be cos(pi/4)-cos(pi/6) instead of cos(pi/6)-cos(pi/4)?
 
No, since:

[tex]\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} sin\phi d\phi = [-cos\phi]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -cos(\frac{\pi}{4}) - (-cos(\frac{\pi}{6})) = cos(\frac{\pi}{6}) - cos(\frac{\pi}{4})[/tex]
 
Last edited:
oh ok...forgot about the negative. thanks!
 

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