Proof that a pyramid of any base shape's volume is \(\frac { 1 }{ 3 } Ah\) <-- yes “Ah”

My History exam just passed (yay) so this is just a small post to (celebrate?). Not that I posted a lot of notes. This is my first one.

Proof that a pyramid of any base shape's volume is $\frac { 1 }{ 3 } Ah$

hi

In order to proof this, we are going to integrate the areas of the cross-section of the solid. Some of you would have already seen this:

$\int _{ b }^{ a }{ A(x)dx }$

Or

$\int _{ b }^{ a }{ A(y)dy }$

Where $A(x)$ is the function for the areas of the cross-section of the solid.

$!$

$\textbf{Finding A(x):}$

First consider this image:

hi

Both triangles are right-angle triangles.

$(h)$ is the height of the solid and $(z)$ is just a constant such that ${ z }^{ 2 }\alpha A$

$(x)$ and $(y)$ are variables which would be used later to define $A(x)$

Since (${ z }^{ 2 }\alpha A$), we can express $A$ as $A=k{z}^{2}$

Where $(k)$ is just another constant. Keep in mind what this constant means, as we would be using it later.

Now, from the image above, it can be seen that both triangles are similar. So, finding the equation of $(y)$ with respect to variable $(x)$ is: $\frac { z }{ h } =\frac { y }{ x }$$y=\frac { xz }{ h }$

And therefore, $A(x)=k{y}^{2}=\frac{k{z}^{2}{x}^{2}}{{h}^{2}}$ The $(k)$ constant is used again to find the equation of the area of the cross-section of the object with respect to variable $(x)$.

$!$ $\textbf{Getting the volume:}$

The volume of the object is: $\int _{ b }^{ a }{ A(x)dx }$$\int _{ 0 }^{ h }{ \frac { k{ z }^{ 2 }{ x }^{ 2 } }{ { h }^{ 2 } } } =\frac { 1 }{ 3 } k{ z }^{ 2 }h$Recall that ($A=k{ z }^{ 2 }$). This makes $\frac { 1 }{ 3 } k{ z }^{ 2 }h=\boxed{\frac { 1 }{ 3 } Ah}$ Since the volume is based on the area of the cross section, the point at the top of the pyramid can literally be anywhere and this everybody-already-knows formula would still work. Thus, the Ah formula has been proved. Hope you liked this note.

$!$$!$

$\textbf{Additional: Volume of Sphere}$

There are 2 ways of doing this. One, is from the method I showed above, which… can be kind of boring….

The second way is a little more interesting.

You can visualise the sphere as made up of infinitely many pyramids. Whose height is the radius of the circle and their collective base area would be the surface area of the sphere. With…more…integration…you…could…find...the…surface…area…of…the…sphere. Which is boring. But we all know it's ($4\pi{r}^{2}$)

So, the volume of the Sphere would be:$\frac { 1 }{ 3 } Ah=\frac { 1 }{ 3 } (4\pi { r }^{ 2 })(r)=\boxed{\frac { 4 }{ 3 } \pi { r }^{ 3 }}$

Ok. This post isn't short.