volume of a tetrahedron??

Question

Use calculus to find the volume (in cubic centimeters) of a tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4 cm, and 5 cm.

Volume = ? cubic centimeeters

Answer 1

consider the base of the tetrahedron at the bottom... it has a leg that's 3 and a leg that is 4.
The area of a given cross section is simply A=(1/2)BH

Note as you move up the tetrahedron's side the triangles get smaller and smaller.

As you are traveling up the tetrahedrons length i guess which is the 5 cm side, if you have infinitely thin cross sections and you add them up along this side you'll have your volume..

So we can construct a general formula for the volume.
V = ∫A(x) dx from 0 to 5

So now all you need is the area of some random cross section in terms of x (the sloping side of the tetrahedron)

when you're at the 0 point along x (the bottom) your sides are a full 3, and 4... As you travel up the sides decrease in proportion to how close you are to the top or (5-x)/5

Assuming you let the base be the 3 sides and the height be the 4 side (doesn't matter).

The base at any point is
3*(5-x)/x

And the height is
4*(5-x)/x

So we can construct A(x)
A = (1/2)BH
A = (1/2) 3 (5-x)/5 4(5-x)/5
A = 6 ((5-x)/5)^2
A = 6/25 (5-x)^2

Now we can setup are integral for volume
V = ∫ 6/25 (5-x)^2 [0 to 5]
V = 6/25 ∫ (5-x)^2 [0 to 5]
V = -6/25 (5-x)^3 (1/3) [0 to 5]
V = -2/25 (5-x)^3 [0 to 5]
V = -2/25 [ 0 - 5^3]
V = -2/25 * -125
V = 250/25
V = 10 cm^3

Answer 2

Well, you know from high school geometry that if the 5 cm edge is the height, the base is a right triangle with sides 3 and 4, area 6 cm², so volume should be (1/3)5(6) = 10 cu cm. Our calculus should give us the same result.

Let the 3 edge be on the x-axis, the 4 edge be on the y-axis, and the 5 edge be on the z-axis. Looking at it in the xz plane, the length of the base shrinks from 3 to 0 as z goes from 0 to 5, so we can write L = 3 - 3z/5. In the yz plane, the width shrinks from 4 to 0 as z goes from 0 to 5, so W = 4 - 4z/5. So we let the integral sum up the areas of a stack of triangles,

V = integral(0 to 5) (1/2)(LW) dz
V = integral(0 to 5) (1/2)(3 - 3z/5)(4 - 4z/5) dz
V = integral(0 to 5) (1/2) [12 - 24z/5 + 12z²/25] dz
V = int(0,5) [6 - 12z/5 + 6z²/25] dz
V = eval(0,5) [ 6z - 6z²/5 + 2z^3/25 ]
V = [6(5) - 6(25)/5 + 2(125)/25] - [0]
V = 30 - 30 + 10 - 0
V = 10 cu cm
just as it should

Answer 3

Perpendicular Faces