Why can't we switch limits of integration for spherical volume calculation?

[ice109]

if i want to find the volume of a sphere why can't i switch the limits of integration on \theta and \phi

i.e. why does this work

\int^a_0 \int^{2\pi}_0 \int^{\pi}_0 r^2 \sin(\theta) d\theta d\phi dr

but this doesn't

\int^a_0 \int^{\pi}_0 \int^{2\pi}_0 r^2 \sin(\theta) d\theta d\phi dr

 

[mathman]

The second integral=0. You are integrating sin over one complete cycle.

 

[maze]

Although both parameterizations cover the whole sphere, the second one turns inside out when it crosses the south pole and goes back up the opposite side.

 

[Defennder]

You might have confused \theta which is the angle on the x-y plane, with \phi which is the azimuthal angle. Some textbooks interpret the notation the other way round. But given the differential volume integral in sphercal coordinates (I'm using the convention stated in the first sentece) which is r^2 \sin \phi dr d\phi d\theta, the limits for \theta go from 0 to 2pi while \phi goes from 0 to pi.

 

[ice109]

The second integral=0. You are integrating sin over one complete cycle.

yes i know

Although both parameterizations cover the whole sphere, the second one turns inside out when it crosses the south pole and goes back up the opposite side.

can you be a little more clear because this pertains precisely to what I'm wondering about

You might have confused \theta which is the angle on the x-y plane, with \phi which is the azimuthal angle. Some textbooks interpret the notation the other way round. But given the differential volume integral in sphercal coordinates (I'm using the convention stated in the first sentece) which is r^2 \sin \phi dr d\phi d\theta, the limits for \theta go from 0 to 2pi while \phi goes from 0 to pi.

no i redefined the angles.

 

[Defennder]

no i redefined the angles.

Well what is your redefinition? It appears in

\int^a_0 \int^{\pi}_0 \int^{2\pi}_0 r^2 \sin(\theta) d\theta d\phi dr

, you have taken the azimuthal angle to be \theta and that should go from 0 to pi. Whereas the other planar angle phi ranges from 0 to 2pi across the x-y plane. So your limits are not correct.

 

[Defennder]

can you be a little more clear because this pertains precisely to what I'm wondering about

He means to say that the azimuthal line which in your incorrect formulation ranges from 0 to 2pi sweeps from the north to south, and then back to the north again. But this covers the same region twice whereas the other planar line associated with phi in your formulation goes from 0 to pi, which doesn't cover the plane. You might want to convince yourself that the volume differential is r^2 \sin \theta drd\theta d\phi by using the Jacobian to evaluate the integral differential variables for dxdydz.