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

A cylindrical hole of radius a is bored through a solid rightcircular 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.
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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,
`(Hh)/H = a/b`
`1h/H = a/b`
`h/H = 1a/b`
`h/H = (ba)/b`
`h = ((ba)/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 (Hh)
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((ba)/b) H`
The volume of small cone is `V_2` ,
`V_2 = 1/3 xx pi xx a^2 xx (Hh)`
`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(ba))/bH  (pia^3)/(3b)H`
`V_3 = (piH)/3(b^2  (3a^2(ba))/b  a^3/b)`
`V_3 = (piH)/3(b^2  a^2/b(3(ba) + a))`
This gives,
`V_3 = (piH)/(3b)(b^3  a^2(3b2a))`
`V_3 = (piH)/(3b)(b^33a^2b+2a^3)`