The discussion revolves around setting up a triple integral over a sphere defined by the inequality x^2+y^2+z^2 ≤ 9, specifically focusing on the integral of the function x^2+y^2+z^2.
There are multiple attempts to establish the correct integral setup, with participants comparing their expressions and questioning the accuracy of the limits and the volume element. Some guidance regarding the volume element has been noted, but no consensus has been reached on the correct integral form.
Participants are exploring the implications of different limits and the arrangement of variables in spherical coordinates, indicating potential confusion about the setup. There is mention of a need to clarify the correct expression for the volume element in spherical coordinates.
PsychonautQQ said:Homework Statement
Triple Integral: x^2+y^2+z^2dV over the ball x^2+y^2+z^2 ≤ 9
Homework Equations
The Attempt at a Solution
so With my integral I had
Triple Integral: p^3sin∅dpd∅dθ
0≥p≥3
0≥∅≥∏
0≥θ≤2∏
Does this look like the correct integral? I swear it is! Yet my answer is wrong. I rebuke these foul math gods!
That looks to me the same as PsychonautQQ posted, just with some of the ≤/≥ turned around the right way and with theta and phi swapped.pasmith said:<br /> \iiint_{r \leq 3} r^2\,dV = \int_0^{2\pi} \int_0^{\pi} \int_0^3 (r^2) (r^2 \sin \theta) \,dr\,d\theta\,d\phi
haruspex said:That looks to me the same as PsychonautQQ posted, just with some of the ≤/≥ turned around the right way and with theta and phi swapped.
Ah yes - the extra r factor.pasmith said:And the correct expression for the volume element ...