Volume of a Square-based Pyramid I've read previous answers that state that the volume of a pyramid is $\frac{1}{3}$ (base $\times$ height). One way to visualize the volume of a square-based pyramid is to envision a cube where every surface is the base of a pyramid. There would be 6 congruent pyramids in the cube and therefore, the volume of any one pyramid would be $\frac{1}{6}$ the volume of the cube. Therefore shouldn't the volume of the pyramid be $\frac{1}{6}$ (base $\times$ height), not $\frac{1}{3}$ (base $\times$ height)? Just a little confused and would appreciate any insight.  
 A: If we call the length of each side of your cube x, then the height of each of the six enclosed pyramids would be x/2. So the volume of each pyramid would be given by:
$$V=\frac{1}{3}\times basearea\times height=\frac{1}{3}\times x^2\times \frac{x}{2}=\frac{x^3}{6}$$
So everything is consistent. :)
A: Six square base pyramids with base $b$ and height $\frac{b}{2}$ can be made into a solid cube of side $b^3$. Hence the volume of each pyramid is $\frac{1}{6} b^3$.
To convert one of these pyramids to a pyramid of height $h$, we need to scale one dimension by $\frac{h}{(\frac{b}{2})}$, and so the volume becomes $\frac{h}{(\frac{b}{2})} \frac{1}{6} b^3 = \frac{1}{3} h b^2$.
A: $$
V
=
\int_{V}{\rm d}V
=
{1 \over 3}\int_{V}\nabla\cdot\vec{r}\ \ {\rm d}V
=
{1 \over 3}\int_{S}\vec{r}\cdot{\rm d}\vec{S}
$$
A: Consider a simple pyramid: this one is square based.

Now we can work out the Area of the base easily but what about the area if we were to cut a slice through it some point as shown above we cut a slice using the plane DE. DE is exactly half way up the pyramid.
At this point the sides of the base are exactly $\frac{1}{2}$ those at the base so the area is  $\frac{1}{4}$.  and of the base at the top is exactly zero.
If we measure the area of the base for any arbitrary slice perpendicular to the base at $x$ where $x$ is the distance from the top, $A_b$ is the area at the base and $h$ is the total height we see.
$$A = \left( \frac{x}{h} \right)^2 \cdot A_b = \frac{x^2}{h^2}  \cdot A_b$$
Here $A$ is the area of the slice at $x$.  Now we can imagine doing this for lots of separate slices at each point multiplying the area of the slice by the slice thickness and adding them up.  The more slices we take the more accurate our answer will be.
With the aid of calculus we can do this with infinitely many  infinitely thin slices to get an exact answer.
$$\begin{align}
V & = \int^{h}_0  \frac{x^2}{h^2} \cdot A_b \ dx \\
  & = \frac{A_b}{h^2} \ \int^{h}_0 x^2 dx \\
  & = \frac{A_b}{h^2} \ \left[ \frac{x^3}{3} \right]^h_{x=0} \\
  & = \frac{1}{3} \ A_b \cdot h
\end{align}$$
