# The large cone has radius 6 cm and height 45.6 cm. (the volume of any cone is given by V = (1/3)π r2h) A frustum is created by removing the smaller cone of height y. Express the volume of the...

The large cone has radius 6 cm and height 45.6 cm. (the volume of any cone is given by V = (1/3)*π* r2h) A frustum is created by **removing** the smaller cone of height y. Express the volume of the frustum as a function of y. (hint: use similar triangles)

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### 2 Answers

Hi, francis,

Since we would be "removing" a smaller cone, then we could subtract a cone where the height would be y. So, volume of the big cone is"

V = (1/3)*π**r^2*h

Of the cone we subtract:

V = (1/3)*π**r^2*y

But, for the smaller cone, r is different. It is smaller than the bigger cone. So, that's what we need to calculate, the radius of the smaller cone.

For that, we draw a cross-section of the cones given on the first attachment. Since the tangent of each triangle would be the same, then the ratio of the sides would be the same. So, we can make:

y/x = h/r

to relate their sides. Solving that for x, the radius of the small cone, we get x = yr/h

So, we then have, for the smaller cone:

V = (1/3)*π**(yr/h)^2*y

Then, subtracting this from the big cone:

V = (1/3)*π**r^2*h - (1/3)*π**(yr/h)^2*y

Factoring out what we can, we get:

V = (1/3)*π*r^2 * (h - y*(y/h)^2 )*

*Now, from there, you can make it into many other forms, like working out the exponent in the parenthesis making y^3. But, I would think this would work. We are given r and h, constants. So, it is in terms of y, the height of the smaller cone.*

*Good luck, francis. I hope this works.*

*Till Then,*

*Steve*

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Oh, sorry, Francis. We could just plug in the radius and height into the formula and simplify the numbers from there. Sorry, I got into "make the formula" mode.

I hope this will still help.

Till Then,

Steve

Smaller cone and larger cone will similar .After removing smaller cone from bigger cone it will form frustum.Let radius of smaller cone is x .

By property of similar triangles

`x/y=6/(45.6)`

`x=(6y)/45.6`

Thus volume of smaller cone is

`v_1=(1/3)pi x^2 y`

`=(1/3)pi ((6y)/45.6)^2y`

`=(36pi)/(3xx45.6)y^3`

`=(12pi)/45.6y^3`

Volume of bigger cone

`v_2=(1/3)pi(6)^2xx45.6`

`=547.2pi`

Thus volume of frustum will be

`v=v_2-v_1`

`=547.2pi-((12pi)/45.6)y^3`

`=(547.2-.263y^3)pi`

Thus volume of the frustum

`v(y)=pi(547.2-0.263y^3)` cubic cm