How would I approach this integration of the volume? Thanks! A cylindrical hole of radius a is bored through a solid right-circular cone of height h and base radius b > a. If the axis of the...

How would I approach this integration of the volume? Thanks!

  1. A cylindrical hole of radius a is bored through a solid right-circular cone of height h and base radius b > a. If the axis of the hole lies along that of the cone, find the volume of the remaining part of the cone. 

Asked on by tcyl

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thilina-g | College Teacher | (Level 1) Educator

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First just draw a cylinder with radius a through the center of the cone without taking the materials out. Now there would be a location at which the cone's radius (Not the base radius, but let's say at a a height of h from the base) be equal to a. Now let's first find this h.

Let the height of the cylinder be H, then,

`(H-h)/H = a/b`

`1-h/H = a/b`

`h/H = 1-a/b`

`h/H = (b-a)/b`

`h = ((b-a)/b)H`

Now if you analyse this colsely you can see that the removed volumes are,

1. A cylinder with radius a and height of h

2. A small cone with base radius a and height of (H-h)

 

The total volume of the cone is V,

`V = 1/3 xx pi xx b^2 xx H`

`V = 1/3pib^2H`

 

The volume of small cylinder is `V_1` ,

`V_1 = pi xx a^2 xx h`

`V_1 = pia^2((b-a)/b) H`

 

The volume of small cone is `V_2` ,

`V_2 = 1/3 xx pi xx a^2 xx (H-h)`

`V_2 = 1/3pia^2xxa/bH`

 

The remaining volume is `V_3` ,

`V_3 = V - V_1 - V_2`

`V_3 = (pib^2)/3H - (pia^2(b-a))/bH - (pia^3)/(3b)H`

`V_3 = (piH)/3(b^2 - (3a^2(b-a))/b - a^3/b)`

`V_3 = (piH)/3(b^2 - a^2/b(3(b-a) + a))`

 This gives,

`V_3 = (piH)/(3b)(b^3 - a^2(3b-2a))`

`V_3 = (piH)/(3b)(b^3-3a^2b+2a^3)`

 

 

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