SOLUTION: Show that the volume of a frustum of a cone = 1/3πh(R²+r²+Rr)

Algebra ->  Test -> SOLUTION: Show that the volume of a frustum of a cone = 1/3πh(R²+r²+Rr)      Log On


   

Question 1210677: Show that the volume of a frustum of a cone = 1/3πh(R²+r²+Rr)
Answer by math_tutor2020(3842) About Me  (Show Source):
You can put this solution on YOUR website!

Here is the side view of a cone.

This is when the circular face is flat on the table. Allowing the cone to point to the ceiling.


Draw a horizontal line segment to form a smaller cone up top.

Removing the top smaller cone will result in a frustum.
A 3D conical frustum resembles a lampshade.
When viewing the lampshade directly from the side, it looks like a 2D trapezoid.

Let's split the diagram in half along a vertical line.


Define the following
R = radius of the larger cone
r = radius of the smaller cone
h = height of the frustum
y = height of the smaller cone
h+y = height of the larger cone

Applying those labels to the diagram gives us this

Let's focus on the right half of the diagram


Based on the last diagram, we have two similar triangles.
We can prove they are similar due to the AA (angle angle) theorem.
Note the shared angles up top, and each triangle is a right triangle.

Because they are similar triangles, the corresponding sides are proportional
It allows us to set up this equation:
r/y = R/(h+y)

Let's solve for y.
r/y = R/(h+y)
r(h+y) = Ry
rh+ry = Ry
rh = Ry-ry
rh = y(R-r)
y = rh/(R-r)
This is the height of the smaller cone.

Volume of a cone = (1/3)*pi*(radius)^2*(height)

Volume of the smaller cone = (1/3)*pi*r^2*y
= (1/3)*pi*r^2*rh/(R-r)
= (1/3)*pi*r^3h/(R-r)

Volume of the larger cone = (1/3)*pi*R^2*(h + y)
= (1/3)*pi*R^2*h + (1/3)*pi*R^2*y
= (1/3)*pi*R^2*h + (1/3)*pi*R^2*rh/(R-r)
= (1/3)*pi*R^2*h( 1 + r/(R-r) )
= (1/3)*pi*R^2*h( (R-r)/(R-r) + r/(R-r) )
= (1/3)*pi*R^2*h*(R-r+r)/(R-r)
= (1/3)*pi*R^2*h*R/(R-r)
= (1/3)*pi*R^3*h/(R-r)

Subtracting the two conical volumes will yield the frustum's volume.

FrustumVolume = LargerConeVolume - SmallerConeVolume
= [ (1/3)*pi*R^3*h/(R-r) ] - [ (1/3)*pi*r^3h/(R-r) ]
= (1/3)*pi*h*( R^3/(R-r) - r^3/(R-r) )
= (1/3)*pi*h*(R^3 - r^3)/(R-r)
= (1/3)*pi*h*(R-r)(R^2+Rr+r^2)/(R-r)
= (1/3)*pi*h*(R^2+Rr+r^2)
= (1/3)*pi*h*(R^2+r^2+Rr)
--------------------------------------------------------------------------

Or a slightly different way is to do this

FrustumVolume = LargerConeVolume - SmallerConeVolume
= [ (1/3)*pi*R^2*h + (1/3)*pi*R^2*rh/(R-r) ] - [ (1/3)*pi*r^3h/(R-r) ]
= (1/3)pi*h ( R^2 + R^2*r/(R-r) - r^3/(R-r) )
= (1/3)pi*h ( R^2 + (R^2*r-r^3)/(R-r) )
= (1/3)pi*h ( R^2 + r(R^2-r^2)/(R-r) )
= (1/3)pi*h ( R^2 + r(R-r)(R+r)/(R-r) )
= (1/3)pi*h ( R^2 + r(R+r) )
= (1/3)*pi*h*(R^2+r^2+Rr)