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Set of all points within a distance of 1 from the box?

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider a solid box with dimensions L,W, and H. Let S be the set of all points whose
    distance is at most 1 from the nearest point inside or on the box. What is the volume of S?


    2. Relevant equations
    Not sure if there are any?


    3. The attempt at a solution
    My initial thought was that S would be a sphere within the box, but after thinking over the problem for a bit, I don't think that's the case. I think my biggest problem here is that I can't imagine the shape of S, so figuring out what to do mathematically is confusing me. I next thought maybe S was a cube within the box, but that doesn't sound right either.
     
  2. jcsd
  3. Sep 7, 2014 #2
    Maybe instead of trying to envision your problem, try to imagine what a lower-dimensional version might look like.

    Given a rectangle, what does the set of points in the plane which are at most 1 from the nearest point inside or on the rectangle look like?

    Or, going down another dimension, given a line segment , what does the set of points in the plane which are at most 1 from the nearest point on the line look like? What does the set of points in 3-space that are at most 1 from the nearest point on the line look like?
     
  4. Sep 7, 2014 #3
    Draw a square, and get a piece of pencil lead. The lead represents the line equaling 1. Have one point of the lead always reside within the square and the other end outside of it. Move the lead around drawing as you go, and you'll begin to see a new shape appear.

    A very similar question would be: given the length of your arms what is the largest area "snow angel" you can create?
     
  5. Sep 7, 2014 #4
    So rather than my initial thought of a sphere inside of the box, is the solution is a sphere outside the box where the distance from the sphere to say, the side of the box, is 1?
     
  6. Sep 7, 2014 #5
    A sphere is something very specific in math, and the object of interest here is not a sphere in general. I suppose one could say that it is a bubble around the box.
     
  7. Sep 7, 2014 #6
    Okay, so I looked up the volume of an ellipsoid and tried to work out the lengths of the axes. If the ellipsoid is 1 unit away from the box, then I figure the lengths of the semi-axes should be 1+.5L, 1+.5W, and 1+.5H, right? And then the volume is V=(4/3)(pi)(1+.5L)(1+.5W)(1+.5H)?
     
  8. Sep 7, 2014 #7
    It's not an ellipsoid either.
     
  9. Sep 7, 2014 #8
    I'm confused then, if it's just a bubble, how can I find the volume?
     
  10. Sep 7, 2014 #9
    Honestly, you're supposed to use your wits. I gave you some advice in my first post that was intended to guide your thinking and help you visualize the problem, but it doesn't seem like you've paid much attention to it.
     
  11. Sep 7, 2014 #10
    I tried drawing it out the way you suggested. For a line segment, I got an oval around the segment that's one unit away from the line at every point. For a rectangle, I got an oval around it, leading me to think that for the 3D box it would be the ellipsoid.

    After looking at ...'s formula, I tried that approach for a rectangle and then a box. I assumed rectangles of L*r and H*r on the sides, with quarter-circles of radius r in the corners, then multiplied by W to account for the third dimension. I got (2Lr+2Hr+(pi)r^2)(W), where r is the radius of the bubble (a maximum of one) and the dimensions of the box are given by L, W, and H.
     
  12. Sep 8, 2014 #11
    I would describe the picture corresponding to the line as "pill-shaped" - straight sides, round ends - and the picture corresponding to the rectangle as a rectangle with rounded corners rather than an oval. Though I suppose that both pictures do resemble some racetracks that are called ovals. The point being that when I think of an oval, as that term pertains to mathematical shapes, I either think of ellipses or of shapes that have some kind of curvature all the way around. All of our shapes are flat in most places*, and I think that's the part that you're struggling to see.

    The shape that your problem is asking about is basically a big rectangular box with rounded edges and corners. The sides are flat. So it's kind of a boxy bubble*.

    *If you make the dimensions small enough - i.e. make the length of the line, the sides of the rectangle, the side lengths of the box very small with respect to the "border" that we're creating - then the resulting pictures do appear to be more circular/spherical/round. But they are not circles/spheres/perfectly round.

    I think what you want to take away from ...'s posts (which are now gone) is that you're meant to chop your solid into "nice" pieces; i.e. parts of circles, rectangles, cylinders, spheres, cubes, etc.
     
  13. Sep 8, 2014 #12

    vela

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    I've attached a drawing of what gopher_p's been trying to get you to draw for shapes in a plane.

    I included a degenerate case where the length of the line segment goes to 0 so that it's just a point. In that case, the resulting shape is a unit circle, and its area is ##\pi##.

    Now imagine from there that you stretch the point out into a line segment of length ##l##. Can you see how the circle has to split into two pieces? What's the total area of the pill-shaped area?

    Now suppose you stretch the line segment vertically into the shape of a rectangle? What happens then and what's the new total area?
     

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  14. Sep 8, 2014 #13
    (delete this post)
     
    Last edited: Sep 8, 2014
  15. Sep 8, 2014 #14
    This is where I've been going wrong. In my last attempt, I sort of envisioned a box with rounded sides of the box but not rounded ends... so the same bubble-box but with flat ends... not really sure why I did that...

    So breaking up the shape, I count one whole cylinder that has the "height" of L (one half on each end of the bubble), one whole cylinder that has the "height" of H (again, one half on each end), and two spheres of radius r (one sphere in all four corners of one end).

    There's also the middle of the box, which, after already talking about the two ends, has a width of W-2, because there's a unit of 1 occupying the radius of the spheres and cylinders, yes? So there's one whole cylinder in the edges of the middle with a height of W-2, two boxes (top and bottom) of volume Lr(W-2) and two boxes (sides) of volume Hr(W-2).

    I believe that accounts for everything, so...

    V = πr2(L+H+(8/3)r+W) + 2r(W-2)(L+H)

    And since the radius is 1, I can simplify to V = π(L+H+(8/3)+W) + 2(W-2)(L+H)

    Have I missed anything?
     
  16. Sep 9, 2014 #15

    haruspex

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    For the external region, you've over counted the spheres. Think about that again, and check your answer by considering the special case where the box is reduced to a single point.
    For the inside, it gets quite complicated in general if you want only points within distance 1 of the surface of the box. But fortunately that is not what the question asks. Read it again.
     
  17. Sep 9, 2014 #16
    So you have a rectangular box and you have a stick. Now how do you place the stick on each box surface to make the far tip of the stick the farthest away from the box? So imagine sticks like that all over each surface sticking out. So that makes up the bulk of the outer edge of the new surrounding surface, but you will notice their are disjoints where each surface of the box connects to the other. So now put the stick at each corner of the rectangular box and move it around and you move it all around until it hits and joins up to the surface made by all the sticks sticking out from all the surfaces. So think if a stick is anchored at one end what sort of shape it sweeps out and then think in this case what fraction of that full shape it could sweep out restricted to fitting within the 1 unit extensions to the other surfaces.

    The volume is the original volume plus the six extended surface pieces (or the three new pairs of extensions) plus 8 connections between each of the new extended main surfaces plus 8 of the corner bits.
     
  18. Sep 9, 2014 #17
    deleted - sorry I misread how the site rendered pi
     
  19. Sep 9, 2014 #18
    deleted - again the pi thing
     
  20. Sep 9, 2014 #19
    You are using too many corners or using too large corners so the 8/3pi term is wrong.

    And the last term 2(W-2)(L+H) seems to be totally confused. I'm not really sure what is even going on there, but it's not like that at all (I mean a couple bits that multiple out of that are OK but that is all the rest are wrong and something is missing).
     
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