Proof that a pyramid of any base shape's volume is 13Ah\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 13Ah\frac { 1 }{ 3 } Ah

hi 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:

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

Or

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

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

!!

Finding A(x):\textbf{Finding A(x):}

First consider this image:

hi hi

Both triangles are right-angle triangles.

(h)(h) is the height of the solid and (z)(z) is just a constant such that z2αA{ z }^{ 2 }\alpha A

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

Since (z2αA{ z }^{ 2 }\alpha A), we can express AA as A=kz2A=k{z}^{2}

Where (k)(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)(y) with respect to variable (x)(x) is: zh=yx \frac { z }{ h } =\frac { y }{ x }y=xzh y=\frac { xz }{ h }

And therefore, A(x)=ky2=kz2x2h2A(x)=k{y}^{2}=\frac{k{z}^{2}{x}^{2}}{{h}^{2}} The (k)(k) constant is used again to find the equation of the area of the cross-section of the object with respect to variable (x)(x).

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

The volume of the object is: baA(x)dx\int _{ b }^{ a }{ A(x)dx }0hkz2x2h2=13kz2h\int _{ 0 }^{ h }{ \frac { k{ z }^{ 2 }{ x }^{ 2 } }{ { h }^{ 2 } } } =\frac { 1 }{ 3 } k{ z }^{ 2 }hRecall that (A=kz2A=k{ z }^{ 2 }). This makes 13kz2h=13Ah\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.

!!!!

Additional: Volume of Sphere\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 (4πr24\pi{r}^{2})

So, the volume of the Sphere would be:13Ah=13(4πr2)(r)=43πr3 \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.

Note by Julian Poon
5 years, 1 month ago

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Holy shit the sphere part was awesome!

Mvs Saketh - 5 years, 1 month ago

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Apple's first note :P Thanks though, I got to know something new :)

Nihar Mahajan - 3 years, 6 months ago

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@Julian Poon Can you add this to a suitable Wiki page? Let me know if you want me to create a skill for you to post to.

Calvin Lin Staff - 5 years, 1 month ago

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ok. but how do you do so?

Julian Poon - 5 years, 1 month ago

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@Calvin Lin I added it to Volume - Pyramid

Julian Poon - 5 years, 1 month ago

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Sorry that the instructions were not clear.

I meant for you to click on "Write a summary", and paste the above text. This way, you will be adding directly to that wiki page, and others can come along and help improve it (like by providing a simplier initial explanation).

Calvin Lin Staff - 5 years, 1 month ago

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@Calvin Lin oh ok. Tomorrow. Its mid-night here :D

Julian Poon - 5 years ago

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https://brilliant.org/discussions/thread/well-this-baffles-me/

Nice! Check out my post.

Danny Kills - 5 years, 1 month ago

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