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:[tex]
\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[/tex]
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 ...
"Integrating over a sphere" refers to the process of calculating the total value of a function over the entire surface of a spherical object.
Integrating over a sphere allows us to calculate important physical quantities, such as the total mass or charge of a spherical object, or the total amount of energy or force acting on the object.
Integrating over a sphere is different because the surface of a sphere is curved, unlike the flat surfaces of other shapes. This requires the use of special mathematical techniques, such as spherical coordinates, to properly calculate the integral.
Some examples of real-world applications include calculating the gravitational force between two planets or stars, determining the total heat transfer in a spherical object, and calculating the total amount of light emitted by a spherical light source.
Yes, integrating over a sphere assumes that the object being integrated over is a perfect sphere, which may not always be the case in real-world situations. In addition, the accuracy of the integration may be affected by the resolution of the grid used to discretize the sphere's surface.