Volume of 3D Shapes

Intro

during Primary school or Highschool, we were taught the volumes of a cone, a pyramid, an a sphere. But where do they come from ? In the textbooks that I used, they proof this volume by using water. Which I am not satisfied.

In this post we will be using Integral to proof the volume of these shapes.

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Basic Concept

assume that we have a function r(x), which can be interpreted as "give me an x-position, and I can give you the radius at that position". The result is the 3D shape created by having our function swiping by rotating r(x) along the X-Axis.

assume that we already magically know the volume of the shape from x_a=0 to x_b=H, let's call that V.

by changing x_b a little bit by dx, the volume will change by dV. Thankfully this change of volume is actually a cylinder, and we know the volume of this cylinder :

by intergrating both sides, we have :

and we can use this for our cone, pyramid, and sphere as well.

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Volume of a Cone

take a right triangle, with a height of H and a width of R, stick the height along the X-Axis and the angle opposite of the width at the origin point of the coordinate system. A cone can be created by rotating the triangle along the X-Axis.

from the Triangle we can know that :

and now we can rearrange this thing and get :

now we can substitute this into the general formula and go from there :

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Volume of a Pyramid

in this case, we cannot exactly rely on r(x), here we can go a dimension higher : A(x) , meaning, "give me an x-position, i'll give you an area".

before diving into proofing the volume, let's take a few steps back.

take a square laying flat on the yz-plane,

if we were to stretch the square along the y-axis by y, the area of the square will increase by y

if we were to stretch the square along the z-axis by z, the area of the square will increase by z

assume we have a 2D shape lay flat on the yz-plane,

this 2D shape can be treated as a bunch of squares (just like what we have earlier)

if we were to stretch the shape along the y-axis by y and z-axis by z

then the area will become :

here, let's create a 2D shape : A_{base}, such that it has an area of 0 when x = 0, and have a depth radius of z and a height radius of y, here we can say when x=H, the depth radius of the base is Z, the height radius of the base is Y , and we can then say :

in this case, we can clarify that Y times Z is not exactly that Area of the shape, however, we can say :

and we can then move on from here :

and now we can start integrating from x=0 to x=H :

and this actually works with any 2D shapes : circle(cone), square(classical pyramid), rectangle(stretched pyramid), polygon, etc

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Volume of a sphere

here we can let one of the tip be the origin of the coordinate system

here we have 2 types of radius :

• R : the main major radius of the sphere

• r(x) : the smaller radius that depends on the x-position

from the Pythagoras theorem, we can know that :

and now we can start integrating from x=0 to x=2R :

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And there we have it, the Volume of a cone, a pyramid, and a sphere. An with the main concept in mind, you can derive a lot of formulas :

"when I change x by dx, the Volume will change by dV, find what is dV, then start integrating"