Calculus I - Volume by Slicing


by GreenPrint
Tags: calculus, slicing, volume
GreenPrint
GreenPrint is offline
#1
Aug11-11, 04:38 AM
P: 1,184
Hi,

I'm trying to solve this problem. "Use the general slicing method to find the volume of the following solids. #13. The tetrahedron (pyramid with four triangular faces), all of whose edges have length 4"

Ok So I don't see how I am suppose to use "the general slicing method" to find the volume of this solid. I had to look up the formula for volume of this solid, sense I didn't know it off the top of my head, V = (sqrt(2)a^3)/12, where a is length of the edge of one of the faces.

So I plugged in 4 for a and got (16sqrt(2))/3. I checked my answer with the back of the book to find that I got the answer correct. I got the right answer but have no idea how to solve this problem using calculus. Thank you for any help you can provide.

If I split the one of the faces, the base, in half so that way it forms a right triangle, given that one of the angles is pi/3, each face is a equilateral triangle, each face has a length of 4, the hypotenuse in this case, the shorter leg is half that 2, the longer leg must be 2sqrt(3) (from c^2=a^2+b^2), the first derivative of the equation that can be used to represent the hypotenuse is sqrt(3), and the equation that can be used to represent the hypotenuse would than be y=-sqrt(3)x+2sqrt(3). I added the negative sign because I don't feel like integrating absolute values.

Solving for x we get
x= 2 - y/sqrt(3)
If I than multiply this equation by two to get the length of the face at any individual point along the hypotenuse (the original equation was only for half the face) I get
x = 4 - (2y)/sqrt(3)

This were I get stuck. I haven't studied geometry in over 4 years, and that was just basic euclidean geometry. I'm unfamiliar with the shape of individual slices that would come out of the base along each point of base. If I knew what shape this is I could just integrate the area formula of it from 0 to 2sqrt(3) assuming the area formula for this shape is some function of height
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HallsofIvy
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#2
Aug11-11, 06:33 AM
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Imagine the tetrahedron sitting on one of its triangular faces. The fourth vertex is connected to the other three. If you slice parallel to the base, you get a triangle that is "similar" to the base. Since the three sides of the base all have length 4, it is an equilateral triangle and so are all slices. You may know, or can calculate, that the area of an equilateral triangle, with side length s, is [itex]\sqrt{3}s^2/4[/itex].

Finally, The side length of the slices depends linearly upon the height. It is 4 when the height, z, is 0 and is 0 at the vertex. You will, of course, need to calculate the height of that vertex. You can, for example, use the Pythagorean theorem to determine the distance from a vertex of an equilateral triangle to its centroid (where the three altitudes cross):
Let the length of a side be "s". Then the length of all altitudes is [itex]\sqrt{3}s/2[/itex]. Call the distance from the centroid to the base of the altitude "a". Then the length of the rest of the altitude is [itex]\sqrt{3}s/2- a[/itex]. And that is the length of the hypotenuse of a right triangle with legs of length a and s/2. The Pythagorean formula gives [itex](\sqrt{3}s/2- a)^2= a^2+ (s/2)^2[/itex]. Solve that for a in terms of s.

Now use the Pythagorean formula again to find the height of the pyramid. The side length of each slice will be kz+ 4, where "k" is chosen to give "0" when z= the height. Integrate the area of each slice, dz, from 0 to that height.


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