Volume of DirtDate: 09/29/97 at 14:38:03 From: Eric Budzynski Subject: Calculus Here's the problem: You have a mound of dirt that is 2 meters high with a bottom radius of 1 meter... Imagine dumping a bucket of dirt onto a table and the shape that it would take. It would sort of be conical in nature but not quite. My problem is to find the volume of this dirt as accurately as possible. I could find the volume of a cone but the shape is not completely accurate to that of a cone. It seems that there may also be a rectangle at the bottom of the shape with a cone on top. I've been told that this problem can be done with calculus. Please help! Eric Date: 09/29/97 at 15:03:24 From: Doctor Rob Subject: Re: Calculus The figure you describe may be the frustum of a cone. Start with a tall cone, and make a horizontal slice, removing a smaller cone from the top. What is left is called a frustum of a cone. You can compute its volume by subtracting the volume of the small cone from the volume of the large cone. To do that, you have to be able to determine the height of the vertex of the large cone. If the large cone has height H and the small cone has height h, and the large cone has base with radius R, the small cone will have base with radius r = R*h/H. The two volumes are then (Pi/3)*R^2*H = (Pi/3)*r^2*H^3/h^2 and (Pi/3)*r^2*h. The difference is V = (Pi/3)*(r^2*h - R^2*H). You know R = 1, so r = h/H, and H = h + 2, and thus H = 2/(1-r) and h = 2*r/(1-r), but you need one more fact to determine r, h, and H exactly. You need the radius of the top, r, for example, or the slope of the sides, H/R = h/r, or some other datum. Lacking that, you have only the formula V = (Pi/3)*2*(1+r+r^2). -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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