Sample Problem 1. Find the volume of the octahedron formed by connecting the centers of the sides of a cube with edge length 1.
At first it seems we have no idea where to start. We don't even know how to find the volume of the if we find one of its side lengths (we need two). Thus, we want to try to think about what information we do have, preferably in 2 dimensions to make it easier to understand. The base (the middle square-shapes part) of the octahedron seems important, so we will look at a cross-section of the plain that it is on. When we do that, will will see that it is simply a square with a smaller square inside it that is formed by connecting the midpoints of the sides of the square (the smaller square is the base of the octahedron). The larger square clearly has side lengths one (as the cross-section was parallel to a face of the cube), so we find that the base of the octahedron has an area of 1/2. Next, we can split the octahedron into two pyramids with the same base and each having a height of 1/2. Using the area we just found of the base, we get each pyramid to have a volume of \(\frac{1}{12}\), so the total volume desired is \(\frac{1}{12}\cdot 2= \frac{1}{6}\).
As you can see in this problem, the key to almost any 3d geometry problem is to split of 3d shapes into smaller 3d shapes and to take 2d cross-sections of important parts of your figure in order to gain needed information, ex. the area of the base. In fact, almost any 3d geometry problem can be split into several simpler 2d geometry problems in order to be solved.
As you can see in this problem, the key to almost any 3d geometry problem is to split of 3d shapes into smaller 3d shapes and to take 2d cross-sections of important parts of your figure in order to gain needed information, ex. the area of the base. In fact, almost any 3d geometry problem can be split into several simpler 2d geometry problems in order to be solved.
As you saw in the last problem, it was important to know the formula for the volume of a tetrahedron. There are several other 3d volume formula that are important to remember listed below. These formulas will not be proved, but you can gain an intuitive understanding of why many of these work by thinking about inscribing them in a cube and comparing the shape's volume to the volume of the empty part of the cube.
- Cube: V=\(\text{area of base}\cdot height=width\cdot height \cdot length\)
- Tetrahedron, cone, pyramid: V=\(\frac{1}{3}(\text{area of base}\cdot height)\)
- Sphere: V=\(\frac{4}{3}\pi r^3\)
Sample Problem 2. Find the volume that the shape below covers when rotated \(360^{\circ}\) around AB.
When we imaging the shape that this forms it looks like two circles, one with radius 3 and one with radius 1, that are 1 unit apart and how their circumferences connected. We don't immediately know how to find the volume of the shape, but it should remind you of a cone. In addition, we know that when you rotate a right triangle around similar to how we rotated the above shape we get a right cone. However, comparing the shape above to a triangle gives us the idea that for some point E on line AB we could draw EA and EC to form right triangles EAC and EBD. In other words, the shape above is just triangle EBD minus triangle EAC for some point E on line AB. However, this would also mean the the 3d shape created by rotating the above shape is just the shape created by rotating triangle EBD minus the shape formed by rotating triangle EAC. However, the shape formed by rotating a triangle is just a cone, and we know how to find the volume of cones! Thus, we just need to find the the distance EA to get the height of both cones. Notice that for E, C, and D to be concentric we would need triangle EAC similar to triangle EBD. Thus, we get \(\frac{3}{2+EA}=\frac{1}{EA}\implies EA=1\). Thus, we get the bigger cone to have height of 3 and the smaller to have height of 1. Therefore, the desired volume is \(\frac{1}{3}\pi\cdot 3^2\cdot 3 - \frac{1}{3}\pi\cdot 1^2\cdot 1=\frac{26\pi}{3}\). The shape that we found the volume of is called a frustum.
The technique used here of thinking of one shape as a larger shape minus a smaller one is important to remember and can be used to express many complicated 3d shapes in terms of simpler ones we are more familiar with.
The technique used here of thinking of one shape as a larger shape minus a smaller one is important to remember and can be used to express many complicated 3d shapes in terms of simpler ones we are more familiar with.
Excercises
- What is the volume of a cube inscribed in a sphere with radius \(\sqrt{3}\).
- What is the volume of the shape formed by connected the centers of each face of a cube with side length of 1?
Solution 1
Note that, by symmetry, the center of the cube is also the center of the sphere. Thus, long diagonal of the cube has length equal to the diameter of the cube, or \(2\sqrt{3}\). Next, we will find the length of the long diagonal of a cube in terms of the cube's side length in order to find the cube's side length.
Note that the long diagonal is the hypotenuse of a right triangle with legs equal to the diagonal of one face and one of the sides. Thus, the square of the length of the long diagonal is equal to to the sum of the square of the length of one side and the square of the length of the diagonal of one face. However, by the Pythagorean Theorem, we also get that the square of the length of the diagonal of one facts is equal to the sum of the square of the length of two sides of the cube.
Thus, the square of the length of the diagonal of the cube is equal to thrice the square of the length of one side. In other words, if d is the long diagonal's length and s is the side length, we get \(d^2=3s^2\implies d=s\sqrt{3}\). Thus, \(s\sqrt{3}=2\sqrt{3}\implies s=2\). Thus, the cube has a volume of 8.
Note that the long diagonal is the hypotenuse of a right triangle with legs equal to the diagonal of one face and one of the sides. Thus, the square of the length of the long diagonal is equal to to the sum of the square of the length of one side and the square of the length of the diagonal of one face. However, by the Pythagorean Theorem, we also get that the square of the length of the diagonal of one facts is equal to the sum of the square of the length of two sides of the cube.
Thus, the square of the length of the diagonal of the cube is equal to thrice the square of the length of one side. In other words, if d is the long diagonal's length and s is the side length, we get \(d^2=3s^2\implies d=s\sqrt{3}\). Thus, \(s\sqrt{3}=2\sqrt{3}\implies s=2\). Thus, the cube has a volume of 8.
Solution 2
Note that the shape formed by connected these centers is just a regular octahedron. To find the volume of the octahedron, we will split it up into two pyramids sharing a base that is on the plane parallel to the base of the cube going through the midpoint of each side perpendicular to the base. We can easily see that this base has an area of half that of the top face of the cube, or an area of 1/2. Because we divided the octahedron into 2, we can easily see that the height of each pyramid is also 1/2. Thus, the volume of the tetrahedron is equal to 2(\(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{3}=\boxed{\frac{1}{6}\)).