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Integrating to find mass

  1. Sep 16, 2012 #1
    Hi i was wondering if i could get help with a question.

    A cone has a radius of 0.75m and a height of 4m. The density changes throughout the cone and can be modeled by the function ρ=3h^2+4h (h for height). Find the mass of the cone.

    Please help
     
  2. jcsd
  3. Sep 16, 2012 #2

    Simon Bridge

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    Welcome to PF;

    There are several ways of doing this - the easiest method is to exploit the symmetry.

    work out how the horizontal radius of the cone varies with height (h) so you have r(h).
    Cut the cone into slices of thickness dh - so the volume of each slice will be ##dV=\pi r^2(h)dh## and so you can find the mass of each slice as a function of h, and then do the integration $$m = \int dm = \int_0^4 f(h)dh$$
     
  4. Sep 16, 2012 #3
    so i'll get
    [o]\int[/4] 3h^4 x π x 9/256 + 4h^3 x π x 9/256 dh
    =96.13kg right?

    the x's are multiplication, still learning how to use the symbols
     
  5. Sep 16, 2012 #4

    Simon Bridge

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    Man that's hard to read ... here, let me help:
    $$ \int_{0}^{4} \left ( 3h^{4}\pi\frac{9}{256} + 4h^{3}\pi\frac{9}{256} \right ) dh$$ ... is that what you meant?

    (If you click the "quote" button at the bottom of this reply, you'll get to see how I did that ;) )

    Frankly I cannot tell if that is right or not - I haven't seen your reasoning, and I don't know which way up the cone goes. However, I was expecting a 4th order polynomial in h so it seems OK.
     
  6. Sep 16, 2012 #5

    HallsofIvy

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    Looking at the cone from positive z axis of an xyz coordinate system, we have two lines, one from (.75, 0) to (0, 4). That has equation y= (-16/3)(x- .75). Solving that for x, the radius of a circle when the line is rotated around the y-axis, x= .75- (3/16)y. The area of a circle of radius x is [itex]\pi x^2= \pi(.75- (3/16)y)^2[/itex] and the volume of a circular disk of that radius and height "dy" is [itex]\pi(.75- (3/16)y)^2dy[/itex]. The mass of such an object with density [itex]3y^2+ 4y[/itex] is [itex]\pi(.75- (3/16)y)^2(3y^2+ 4y)dy[/itex]. To find the total mass, integrate from y= 0 to y= 4.
     
  7. Sep 16, 2012 #6

    Simon Bridge

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    Why can't the lines be from (0,0) to (0.75,4) etc. r(h)=3h/16 in that case.
    It is quite common to define cones like that, though I'd normally think of the point as "higher" than the wide part. Hopefully OP knows which way up the cone goes ;) In fact - that seems to be the way around OP did it.
     
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