Another thread about integration!

titaniumx3

titaniumx3

titaniumx3

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I'm really confused on this question (no seriously I am!) and would really appreciate some input.

A solid hemisphere H of radius a has density depending on the distance p from the centre of the base disc and given by 2a-p. Write down the mass of the hemisphere as a triple integral in: spherical polar, cylindrical polar and cartesian coordinates.
Click to expand...

Now, I have no problem creating the triple integrals in the three coordinate systems but what confuses me is the density and mass functions. For example in the cartesian form, would the integral simply be:

int(int(int(2*a-z))).dz.dy.dx ?

Does the p simply mean movement along the z-axis or am I missing something? Will the same thing apply to the other coordinate systems? Sorry if this is a really silly question, but I really appreciate some clarification on this. :confused:
 
titaniumx3 said:
I'm really confused on this question (no seriously I am!) and would really appreciate some input.



Now, I have no problem creating the triple integrals in the three coordinate systems but what confuses me is the density and mass functions. For example in the cartesian form, would the integral simply be:

int(int(int(2*a-z))).dz.dy.dx ?

Does the p simply mean movement along the z-axis or am I missing something? Will the same thing apply to the other coordinate systems? Sorry if this is a really silly question, but I really appreciate some clarification on this. :confused:
Click to expand...

Well, for the mass of an object, you have:

M = int int int (density) dV

Where the integrals are taken over the boundaries of the body in question.

The density you've been given is in spherical co-ordinates.

So express dV in terms of rho, phi and r, and the denisty function likewise, and intergrate. DO similar for cartesian and cylindrical cases. In each case, pay attention to the transformation of the boundary of the object into the new co-oprdinate system.
 
The mass M of an object with density f is given by

M = ʃʃʃ f dV

where dV is an appropriate expression for the coordinate system that you're using. For example, in cartesians you have dV = dx dy dz, in cylindrical polars you have dV = r dr dθ dz and in spherical polars you have dV = r² sinθ dr dθ dФ (where Ф is the angle 'around' the vertical axis, and θ is the azimuthal angle, i.e. the angle from top to bottom of a sphere). It's then slightly complicated by the fact that you have to use the right limits.

Here your density is given by f = 2a - r, so that you have density 2a at the centre of the hemisphere (i.e. at the middle of the base) and density a at the edges.

The easiest way to do this is to just calculate the mass of a sphere, and then halve it. I'll do it in spherical polars:

M = ʃʃʃ (2a - r) r² sinθ dr dθ dФ

... = 4π ʃ (2a - r) r² dr

where I've just done the integrals in θ and Ф (they're easy) with the appropriate limits. Then doing the r integral, you should get

M = 5πa^4/3

And don't forget to halve it, because you want the mass of a hemisphere, not a sphere. Does that make sense?
 
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That makes a lot more sense! :) So the 2a-p density function is given in spherical coordinates. Hence when writing it in cylindrical and cartesian coordinate systems, would I be correct in using 2a-sqrt(r^2+z^2) and 2a-sqrt(x^2+y^2+z^2) respectively?
 
titaniumx3 said:
Isn't that supposed to be M = 4πa^4/3 ?
Click to expand...
Possibly - I have been known to make the odd algebraic mistake, especially late at night.

You're completely right in that you'd write sqrt(r^2 + z^2) for the radial distance in cylindrical polars.
 
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