Workspace Volume in Spherical Coordinates

In summary, the conversation is about a problem with calculating the volume of a workspace using spherical coordinates. The integral setup and actual volume given by a CAD program are discussed, along with confusion over the notation for theta and phi and the orientation of the coordinate axes in the diagram. The importance of the order of integration in the integral and the derivation of the differential element in spherical coordinates is also mentioned.
  • #1
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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)

[tex]\rho[/tex]: 1 to 1.1 meters
[tex]\theta[/tex]: 0.05 radians
[tex]\phi[/tex]: 0.1 radians

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

Volume = [tex]\int^{1.1}_{1}\int^{0.05}_{0}\int^{0.1}_{0} \rho^2sin\theta \delta\phi \delta\theta \delta\rho[/tex]

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~
 

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

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.
 
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  • #3
swraman said:
Which axes does theta and phi revolve around?
The reason I ask is because in common notation, the Jacobian (the r^2sin() ) ) is [tex]r^{2}sin( \phi)[/tex] and not [tex]r^{2}sin( \theta)[/tex]

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 [tex]\phi [/tex] and [tex]\theta [/tex].

If a sphere is uniform all around, does it really matter whether we integrate [tex]\phi [/tex] first or [tex]\theta [/tex] 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.
 
  • #4
hi_hopes said:
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 [tex]\phi [/tex] and [tex]\theta [/tex].

If a sphere is uniform all around, does it really matter whether we integrate [tex]\phi [/tex] first or [tex]\theta [/tex] 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 [tex]\rho ^{2} sin(\phi)[/tex], 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. [tex] dV = \rho ^{2} sin(\phi ) d\rho d\phi d\theta [/tex]
 

What is workspace volume in spherical coordinates?

Workspace volume in spherical coordinates refers to the amount of space that can be occupied by an object or system within a spherical coordinate system. It is a measurement of the maximum distance an object can move in each direction from a central point in spherical coordinates.

How is workspace volume calculated in spherical coordinates?

The workspace volume in spherical coordinates is calculated using the formula V = 8/3πr^3, where r is the radius of the sphere. This formula takes into account the three dimensions of space in spherical coordinates.

Why is workspace volume important in scientific research?

Workspace volume is important in scientific research because it allows researchers to determine the limitations and capabilities of an object or system within a spherical coordinate system. It also helps in designing and optimizing systems for maximum efficiency and effectiveness.

What factors can affect the workspace volume in spherical coordinates?

The workspace volume in spherical coordinates can be affected by the radius of the sphere, the shape and size of the object or system, and any constraints or boundaries within the coordinate system. Other factors such as external forces or obstacles can also impact the workspace volume.

How can the workspace volume in spherical coordinates be visualized?

The workspace volume in spherical coordinates can be visualized using computer-aided design (CAD) software or by creating physical models. These tools allow researchers to see how an object or system can move within the coordinate system and identify any limitations or challenges.

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