What are the Lines of Symmetry in a Tetrahedron?

[endeavor]

Find the volume of "A tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4 cm, and 5cm."

This is how I visualized it:
http://img282.imageshack.us/img282/9466/calculus31re.th.jpg

The area of a triangle along the x-axis is:
A(x) = .5 * x * (3/5x), 3/5x is from similar triangles.
A(x) = 3/10 * x²
$$V = \int^{5}_{0} A(x) dx = \frac{3}{10} \int^{5}_{0} x^2 dx<br /> = \frac{1}{10} [x^3]^{5}_{0}<br /> = 12.5 cm^3$$
But the answer is 10 cm³. Why doesn't my method work?

 

[StatusX]

Right now your dx is wrong. It corresponds to a change in the length of the short side of the triangluar slice, where as it should be the thickness of the slice, which is dy. You need to find dx/dy, the change in the length of this side for a change in height y.

 

[benorin]

The area of a right triangle is A=b*h/2, where b=base, h=height
In your diagram, slice through your tetrahedron with a plane passing through a given x value (the shape made by this is a right triangle whose right angle has its corner on the x-axis). The height of that triangle is
h=4(1-x/5) since the line that connects the vetricies on the y-axis and the x-axis is y/4+x/5=1; the base of that right triangle is b=3(1-x/5), since the line that connects the vetricies on the x-axis and the z-axis is
z/3+x/5=1; hence the cross-sectional area is A(x)=3(1-x/5)*4(1-x/5)/2
simplified this gives A(x)=6(1-x/5)^2, so the volume is

$$6\int_{0}^{5}\left( 1-\frac{x}{5}\right) ^2dx =10$$

 

[finchie_88]

Would it not be easier to do that using a triple scalar product ( $V = \frac{1}{6}a \cdot (b \times c)$ ), where a, b and c are the vectors of the sides?

 

[endeavor]

The area of a right triangle is A=b*h/2, where b=base, h=height
In your diagram, slice through your tetrahedron with a plane passing through a given x value (the shape made by this is a right triangle whose right angle has its corner on the x-axis). The height of that triangle is
h=4(1-x/5) since the line that connects the vetricies on the y-axis and the x-axis is y/4+x/5=1; the base of that right triangle is b=3(1-x/5), since the line that connects the vetricies on the x-axis and the z-axis is
z/3+x/5=1; hence the cross-sectional area is A(x)=3(1-x/5)*4(1-x/5)/2
simplified this gives A(x)=6(1-x/5)^2, so the volume is

$$6\int_{0}^{5}\left( 1-\frac{x}{5}\right) ^2dx =10$$

This works...
If I used a triangle perpendicular to the y-axis (along the x and z axes), then it seems like I would have to find the area in terms of y, right? But why doesn't finding the area in terms of x work?

 

[benorin]

x is never the height nor base of any right triangle form by slicing that tetra hedron by a plane parallell to one of the coordinate planes.

 

[Tide]

You can check your answer by noting that the volume of a pyramid is 1/3 the area of its base times its height.

 

[Lamoid]

The area of a right triangle is A=b*h/2, where b=base, h=height
In your diagram, slice through your tetrahedron with a plane passing through a given x value (the shape made by this is a right triangle whose right angle has its corner on the x-axis). The height of that triangle is
h=4(1-x/5) since the line that connects the vetricies on the y-axis and the x-axis is y/4+x/5=1; the base of that right triangle is b=3(1-x/5), since the line that connects the vetricies on the x-axis and the z-axis is
z/3+x/5=1; hence the cross-sectional area is A(x)=3(1-x/5)*4(1-x/5)/2
simplified this gives A(x)=6(1-x/5)^2, so the volume is

$$6\int_{0}^{5}\left( 1-\frac{x}{5}\right) ^2dx =10$$

Sorry to resurrect this old thread, but I can't figure out why the line y/4+x/5=1 or the line z/3+x/5=1 are equal to 1?