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    Ok so there is part of the plane x+y+z=1 which lies in first octant.

    I was meant to find the area - which i did by projecting on x,y plane and working out double integral to be

    root3 /2 (which i hope is right!)

    then it asks to find its centre of mass..and use geometrical arguments to check the result for the area..

    how do i do these two parts?
    thanks..

    (im guessing the geometrical argument is just that it is and eighth of the area of some figure? im not sure )
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    shameless *bump*
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    (Original post by King of TSR)
    Ok so there is part of the plane x+y+z=1 which lies in first octant.

    I was meant to find the area - which i did by projecting on x,y plane and working out double integral to be

    root3 /2 (which i hope is right!)

    then it asks to find its centre of mass..and use geometrical arguments to check the result for the area..

    how do i do these two parts?
    thanks..

    (im guessing the geometrical argument is just that it is and eighth of the area of some figure? im not sure )
    Center of mass is figured out by finding out the center of mass for all three of the dimensions (x/y/z) and combining that into a coordinate point.

    The way you find any one of those is 'weighting' the mass of the solid by its distance from either the x, y, or z axis and then dividing by the mass of the entire figure.
    Basically your integral for the center of mass with respect to x should look something similar to

     \frac{\int\int\int{x*\delta}dxdy  dz}{\int\int\int{ \delta}dxdydz} where your limits of integration define the solid in question, the area below x+y+z=1 in the first octant. \delta is the density function of the solid which can be assumed to be any constant (generally 1) if it's not stated - like in this question.

    To explain that integral further, the top part represents the 'weighted' value of the masses in the x-direction and when divided by the mass of the figure as a whole (found by integrating the density function over the solid), you get the x-coordinate of the center of mass.

    To figure out the y and z components, just replace the 'x' in that original integral with a 'y' or a 'z' respectively.

    ---------

    As for the geometric part, since the figure has a constant density (generally assumed to be one for simplicity's sake), you can look at it logically and see where the center of mass would be. The center of mass would be at the point that splits the mass of each dimension in half.

    For example, if you have a cube of constant density (which you don't here but it's the same general idea), the center of mass will be the place that splits the figure's mass in half in each dimension. For example, in this cube, if the cube was drawn between x = 0 to x = 4, y 0 to y = 4, and z = 0 to z = 4, the center of mass would be at (2,2,2) because it splits the cube's mass in half in each dimension.

    -- I'm sure my second explaination isn't too good. I'm quite tired now but I'll try to rexplain it in the morning. I would say a good first step is to go through and do it mathematically then geometrically.
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    (Original post by King of TSR)
    then it asks to find its centre of mass..and use geometrical arguments to check the result for the area..
    Is the figure not just an equilateral triangle with vertices at the unit vectors along each axis? This should make it easy to find its centre of mass, and evaluate the area in a geometric argument.
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    (Original post by GHOSH-5)
    Is the figure not just an equilateral triangle with vertices at the unit vectors along each axis? This should make it easy to find its centre of mass, and evaluate the area in a geometric argument.

    Yes it is!

    So, yes, the geometrical argument is quite easy..how do I do it the integral way though? (the q wants both methods.)
    thanks
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    (Original post by King of TSR)
    Yes it is!

    So, yes, the geometrical argument is quite easy..how do I do it the integral way though? (the q wants both methods.)
    thanks
    I'm afraid I'm not sure how to evaluate the area with integration; although, one could rotate (transform) the coordinates so that the vector (1,1,1) is mapped to be parallel to the vector (0,0,1), so that our plane in question is now 'flat', and then it just becomes integrating the area between appropriate lines.
 
 
 
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