volume using spherical cordinates

[aron101782]

a wedge from a unit sphere pi/6 off the x axis


I(0, pi/6)I(0,pi)I(0,1) p^2sinb dp da db
pi/6 1/3 2 = pi/9


a wedge from a unit sphere pi/6 off the z axis
I(0, pi)I(0,pi/6)I(0,1) p^2sinb dp da db
pi/3 (1-3^1/2 /2)


why question is why are these two wedges not the same. It would appear to me that simply taking the cut from the z axis wouldn't change anything. Although that must not be the case given the calculations and differences in the sketches made by Mathematica. If somebody could please help me to conceptually visualize this case.

 

[aron101782]

my mistake I made a typo in the euqations

I(0, pi/6)I(0,pi)I(0,1) p^2sinb dp db da

I(0, pi)I(0,pi/6)I(0,1) p^2sinb dp db da

it would appear as if the second is the same as the first only rotated to the right 90 degrees such that the positive x axis replaces the positive z axis why do the calulations and sketches yield different results. I suppose it may not be exactly the same since that when simply rotating the slice for pi cuts down the same line as oppse to swapping the limits of integration results in the pi slice along a different line.

 

[aron101782]

A good question is that if in fact those regions are not identical then how would it be possible to take that same cut from the top of the sphere using sherical cordinates

 

[LCKurtz]

I'm guessing you aren't getting much response for two reasons. First, you are using non-standard notation with you p,a,b instead of the usual [MATH]\rho,~\phi,~\theta[/MATH] and not explaining which is which. Secondly, you aren't using tex so your "integrals" are very difficult to read. And the reason you are getting different answers is the regions don't have the same shape. And one of your graphs is wrong but it's hard to say which since neither is scaled properly nor labeled.

 

[aron101782]

then how would you calculate I(0, pi/6)I(0,pi)I(0,1) p^2sinb dp db da if it were simply rotated 90 degrees such that the angle off the zaxis would be pi/6. The only way I see to get to that wedge would also be I(0, pi)I(0,pi/6)I(0,1) p^2sinb dp db da. the only difference is that theta slices horizontaly.
such a wedge exists and if that doesn't get to it what does in spherical coordinates. Its simply the cut out of the sphere that pi/6 off the diameter. Im no good with sketches or mathemtica to really try and illustrate what I am visualizing.

 

[aron101782]

I have tried to illustrate this on a baseball and it would appear that I(0, pi/6)I(0,pi)I(0,1) p^2sinb dp db da , wraps around half of the ball

while I(0, pi)I(0,pi/6)I(0,1) p^2sinb dp db da , theta appears to be bound by pi/6.

see this video: