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Finding mass from density: please help

  1. Apr 12, 2006 #1
    Hi,

    Today at the bathroom (an atmosphere I find to be highly conductive to the exercise of grey matters) I thought up the following problem and mentally solved it in the simpler but longer method, but ran aground when I tried to do it in the shorter method. First I state the problem and how I solved it.

    Q: A cube with sides of 1 meter is built in such a way that in the rectangular Cartesian coordinate system built by taking a certain corner O as the origin and the three sides meeting at that vertex as the x, y and z axis, the density of the cube at a point (x, y, z) is given by f(x, y, z)=(x+y+z) kg/m^3. Find the mass of the cube.

    My method:

    Let us consider a particular straight line in the cube found by holding x and y constant. Taking the scale of the coordinate system to be 1 meter, we can see that z can vary between 0 and 1. Let us consider the mass of the infinitesimal portion of this line lying between z and dz. Clearly this mass [tex]dM_1 = (x+y+z)dz[/tex] (because mass = density x volume). Therefore the mass of this complete straight line is given by

    [tex]\int dM_1 = \int_0^1 (x+y+z)dz = \left[ (x+y)z+\frac {1}{2}z^2 \right]_0^1 = (x+y)+\frac{1}{2}[/tex]

    Now we consider this mass to be localised at a point (assume the z=0 endpoint for simplicity) of the straight lines, effectively eliminating the z dimension from the calculation. Visualise squeezing all the straight lines standing on the xy-plane onto their end-points, turning the cube into a plane. We proceed to apply this process twice more.

    Consider the straight line formed by holding x constant (it will not be a plane anymore because the dimension has been reduced). Like before, the infinitesimal mass is [tex]dM_2 = (x+y+\frac{1}{2})dy[/tex] (the density has changed because the mass instead of being distributed over the cube is now localised at the points of the xy-plane). The mass of the straight line

    [tex]\int dM_2 = \int_0^1 (x+y+\frac{1}{2})dy = \left[ (x+\frac{1}{2})y+\frac{1}{2}y^2 \right]_0^1 = x+1[/tex]

    As before, consider this mass to be localised at the point y=0 of each straight line, eliminating another dimension and turning the plane into one straight line (visualise squeezing the plane onto one of its sides). We apply this process one last time.

    Take the straight line. The infinitesimal mass is [tex]dM_3 = (x+1)dx[/tex]. The mass of the straight line

    [tex]\int dM_3 = \int_0^1 (x+1)dx = \left[ x+\frac{1}{2}x^2 \right]_0^1 = 1+\frac{1}{2} = \frac{3}{2}[/tex]

    As the mass of the whole cube is equal to this single straight line (the line having been formed by repeatedly localising the mass of the cube to one lower dimension), we have the mass of the cube as 1.5 kg, unless of course I've committed a dreadful mistake somewhere. We shall choose to neglect the quantum-relativistic corrections originating in such calamities for now :)

    As anyone can see, this is a dreadfully long procedure (so long in fact that I've a suspicion no one is reading this right now), so I attempted something shorter.

    Let all the three coordinates vary by the infinitesimal quantities dx, dy and dz from x, y and z respectively. The mass of the resulting portion is given by [tex]dM = (x+y+z)dxdydz[/tex]. Integrating this 3-form over the entire cube (represented by the letter C here) we have [tex]\int dM = \int_C (x+y+z)dxdydz[/tex]. Our earlier calculations suggest this integral should equal 1.5.

    Therefore my question (at last) is how can I evaluate the integral given above? I'm only just starting on my 11th grade, so please don't send things over my head. Thanks a lot in advance.

    Molu
     
    Last edited: Apr 12, 2006
  2. jcsd
  3. Apr 12, 2006 #2

    arildno

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    What you have basically shown, is called Fubini's theorem:
    A triple integral can be evaluated by REPEATED INTEGRATION, i.e, first integrate with respect to z (x and y considered constants), then with respect to y (x considered constant), and finally with respect to x.
     
  4. Apr 12, 2006 #3

    nrqed

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    Thanks for the laughs :biggrin:

    What you did is right and it is *excatly the same* as calculating [tex]\int dM = \int_C (x+y+z)dxdydz[/tex].

    This can be written as the sum of three integrals. Consider the first one. It is [itex]\int_C x\,dxdydz= \int_0^1 dx\int_0^1 dy \int_0^1 dz \,x = {x^2 \over 2} \vert_0^1 y \vert_0^1 z \vert_0^1 = {1 \over 2} [/itex] and you get also 1/2 and 1/2 for the other two for a total of 3/2. but this is completely equivalent to what you did.

    Except people prefer to solve those probelmes by saying mass= integral over density over a volume, or [itex] m= \int dV \rho [/itex] which will work for any type of mass distribution or object shape (although in genetal the integrals are really tough).

    Patrick
     
  5. Apr 12, 2006 #4
    But how do I evaluate the integral of the 3-form directly? Incidentally, I thought triple integral was integrating thrice consecutively with respect to the same variable (like third-order derivative).
     
  6. Apr 12, 2006 #5

    arildno

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    It is NOT exactly the same, nrged, unless you make use of Fubini's theorem, by which repeated integration is proven to get the same result as, say, taking the limit of partial 3-D Riemann sums.
     
  7. Apr 12, 2006 #6

    arildno

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    Well, you could do it the extra-ordinarily hard way of calculating the limit of 3-D Riemann sums..

    Alternatively (as everyone else does), use repeated integration.
     
  8. Apr 12, 2006 #7
    It's a relief to know I was right, this is the first time I attempted a problem in multivariate calculus! I didn't expect to be able to get it right in 11th grade, but it sort of flowed out into the keyboard :) Could you explain that last equation a little? How would you integrate OVER density OVER volume. I mean, if it's all OVER then what ARE you integrating? Can you please give a simple example?
     
  9. Apr 12, 2006 #8
    Sounds horribly like something right out of HP LOvecraft :D
     
  10. Apr 12, 2006 #9

    nrqed

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    What I meant is that the *mathematical steps* he went through are exactly the same as evaluating the 3-D integral (he could do the dz then the dy then the dx and the mathematical steps are exactly the same as doing the 3-d integral he gives at the end). But I see what you are saying, the interpretation is different. I was talking about the mathematical steps. Sorry for the confusion.
     
  11. Apr 12, 2006 #10

    nrqed

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    unfortunately I have to go teach.

    But, in a nutshell: you are summing (infinitesimal) masses. Think of it this way: The mass density if rho (x+y+z in your example). Consider a little cube of sides dx, dy and dz. It has a volume dx dy dz. The mass of this infinitesimal cube is rho dxdydz. Now you sum it up over all the tiny cubes making up your entire volume. SO you are adding *masses*. But you are integrating over x, y and z and the function rho(x,y,z) "calculates" for you the mass of each tiny cube. We have to take tiny cubes so that the function rho is pretty much constant over the volume of the tiny cube (in other words, we take dx and dy and dz small enough that evaluating rho at x,y,z or x+dx,y,z or x,y+dy, z+dz, and so on does not make any difference. Ofc ourse all of this can be made rigorous using calculus.
     
  12. Apr 12, 2006 #11
    I'm not sure I understand this. No hurry, answer when you come back, but you just described my process. I was asking about the last equation you posted, the one you said most people use but the integrals are very tough. What was that and can you give an example please? Thanks.

    Another question I have is about the rigour. I know little of whatever theory there may be in this matter, I did the calculation intuitively from having studied the elementary single-variable calculus textbook in 9th grade in my spare time (we are taught calculus towards the end of 11th grade and in 12th grade) and having read mass = density integrated over volume in a popular science book, so are there any unstated assumptions or similar loopholes in the calculation?
     
    Last edited: Apr 12, 2006
  13. Apr 12, 2006 #12
    I get it, like the Fundamental Theorem of Calculus, right? Antiderivatives (aka Indefinite Integrals) are conceptually distinct from the Definite Integrals but are shown to be equivalent in the theorem.
     
  14. Apr 12, 2006 #13

    nrqed

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    Reading this again more carefully, I want to point out that stricttly speaking, this is incorrect. If you mutiply rho by a distance, you do not get a mass, but a surface mass density (kg/m^2). If you think of a very thin rod with square cross section of sides dx and dy instead, then after multiplying the above by dx dy you get the mass of this thin rod (of length 1 along z and of sides dx and dy).
    ? But you integrated over z so you cannot bring it back to a "point". You cannot fix z to be zero! You have integrated from z=0 to z=1!
    I think you are thinking about a "point" in the xy plane. But again, it's not really apoint, it's a little square or sides dx and dy (the cross section of the rod)
    You are not "turning the cube into a plane", you are adding stacking up a bunch or rods up from y=0 to y=1, all with sides dx and dy. That produces a plane which is a section of the whole cube.

    This is not quite logical. The entire mass is distributed over the entire cube. I am not sure what you mean here.


    I had looked at the maths (the integrals) but when I looked at your explanations, there seemed to be several conceptual problems.


    Patrick
     
  15. Apr 12, 2006 #14
    Actually, that was just a visualisation. I'm imagining for the sake of calculation that the mass of the whole lien is concentrated onto its endpoint, visually equivalent to squeezing the cube onto a surface. The integration is giving me the mass (or the more accurate surface mass density) of the whole straight line, but for the purpose of computing the mass it can be taken that the whole of the mass is localized onto one point. We are not literally squeezing the cube, it's simply that the cube and the square are equivalent for the purpose of computing the mass.

    Once again, we are not literally redistributing the mass, just simplifying the calculation by transforming the mathematical model. We are transforming the situation into something that is mathematically equivalent to the original model for the purposes of this calculation.

    As for the matter of rho times length not being mass, that was a mistake of terminology on my part. It is not mass. As we are reducing the dimension by one and localising the mass of the cube onto one of its surfaces, the original mass density (kg/m^3) gets converted into a surface mass density (kg/m^2), which is then converted into a length mass density (kg/m) and then finally into the mass.
     
    Last edited: Apr 12, 2006
  16. Apr 12, 2006 #15
    One last question, is the explicit expression for the mass density distribution invariant under GL or any subgroup of it? To summarise the questions,

    1)to nrged: Can you explain the last equation you wrote?
    2)What are the loopholes in the calculation?
    3)Does the explicit expression of the density in coordinates obey any symmetry?

    Thanks.
     
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