Workspace Volume in Spherical Coordinates

[hi_hopes]

Hey all,

I'm having trouble calculating a "workspace" volume.

The volume is easiest modeled by using spherical co-ordinates, and is defined by these boundaries:

(assume variables r, theta, phi)

$$\rho$$: 1 to 1.1 meters
$$\theta$$: 0.05 radians
$$\phi$$: 0.1 radians

Here's how I've set up my integral:

Volume = $$\int^{1.1}_{1}\int^{0.05}_{0}\int^{0.1}_{0} \rho^2sin\theta \delta\phi \delta\theta \delta\rho$$

I've used Pro/ENGINEER to give me the actual volume of this section: ~0.000595 m^3
But..that's not even close to what I'm getting when I solve that integral!

Anyone see where my problem is?

Thanks in advance~

 

[swraman]

Which axes does theta and phi revolve around?
The reason I ask is because in common notation, the Jacobian (the r^2sin() ) ) is $$r^{2}sin( \phi)$$ and not $$r^{2}sin( \theta)$$

Also, the picture you attached doesn't look like it goes from the limits of phi that you specified, although thyis could be because of the perspective t is drawn from. is it possible to show the problem with the coordinate axis in the diagram?

Using the notation I've always used for spherical coordinates, I get ~2.7560e-005 as the solution.

 

[hi_hopes]

Which axes does theta and phi revolve around?
The reason I ask is because in common notation, the Jacobian (the r^2sin() ) ) is $$r^{2}sin( \phi)$$ and not $$r^{2}sin( \theta)$$

Also, the picture you attached doesn't look like it goes from the limits of phi that you specified, although thyis could be because of the perspective t is drawn from. is it possible to show the problem with the coordinate axis in the diagram?

Using the notation I've always used for spherical coordinates, I get ~2.7560e-005 as the solution.

Thanks for your reply :)

I may have come up with your result once or twice...but it is still less than half of what the CAD program gives me...

To be honest, I'm not familiar with the standard notation of $$\phi$$ and $$\theta$$.

If a sphere is uniform all around, does it really matter whether we integrate $$\phi$$ first or $$\theta$$ first - or what axis they revolve around? Basically, the shape is twice as long in one of the angular dimensions as the other.

The image is from a random perspective - I only put it there so others can visualize the part.

 

[swraman]

Thanks for your reply :)

I may have come up with your result once or twice...but it is still less than half of what the CAD program gives me...

To be honest, I'm not familiar with the standard notation of $$\phi$$ and $$\theta$$.

If a sphere is uniform all around, does it really matter whether we integrate $$\phi$$ first or $$\theta$$ first - or what axis they revolve around? Basically, the shape is twice as long in one of the angular dimensions as the other.

The image is from a random perspective - I only put it there so others can visualize the part.

It doesn't matter which order you integrate them in, but since the integral is $$\rho ^{2} sin(\phi)$$, it does make a difference which limit phi is and which limit theta is. This comes up because of the derivation of the differential element dV in spherical corinates. $$dV = \rho ^{2} sin(\phi ) d\rho d\phi d\theta$$