# Find the volume of a right circular cone with height 18 and base radius 6. Find the volume using integrals. Do it the way volume is found in cal 2

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### 1 Answer

You need to consider a cone, centered on y axis, whose top is in origin.

You should use the following formula to find the volume of the cone using integral such that:

`V = int_a^b A(x) dx`

Notice that the cross section of the cone is a triangle, whose base is 2R and height is h.

Using similar triangles yields the following ratios such that:

`R/h = r/(h-x) => r = (R/h)(h-x)`

Evaluating the area of the cross section yields:

`A(x) = pi*r^2 => A(x) = pi*(R/h)^2(h-x)^2`

You should substitute `pi*(R/h)^2(h-x)^2` for `A(x)` in formula of volume such that:

`V = int_0^hpi*(R/h)^2(h-x)^2 dx = pi*(R/h)^2*int_0^h(h-x)^2 dx `

You should come up with the following substitution `h-x = t => -dx = dt => dx = -dt` `V = -pi*(R/h)^2*int_h^0 t^2dt => V = -pi*(R/h)^2*(t^3/3)|_h^0`

`V = -pi*(R/h)^2*(-h^3/3)`

`V = (pi*R^2*h)/3`

Substituting 18 for h and 6 for R yields:

`V = pi*36*18/3 => V = 216 pi`

**Hence, evaluating the volume of cone, using integral, yields `V = 216 pi.` **

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