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loom91
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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
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
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