# calculate the volume of water in a portion of a cone

Imagine we have a cone filled with water, if we were to take the upper portion of that cone how would we calculate the volume of water present. For example:

So, in this example we have a surface area of 100 m^2, a depth of 300 m and a length at the base of 10 m. Is this solved by calculating the area at the bottom of the cone i.e. pi*0.5*10 ^2 and then when combined with the depth we could calculate the volume?

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It is called frustum of a cone. – Maesumi Jan 10 '13 at 20:08

I would calculate the volume of water in the entire cone, and then subtract off the volume in the truncated portion. This requires you to figure out the depth of the entire cone. You can do this by considering a right triangle having a "central" leg through the center of the cone, a "top" leg across the surface, and the hypotenuse running down the side of the cone:

        y
________
|      /
300 |     /
|_10_/
|   /
x |  /
| /
|/


You can figure out $y$ using the given area of the surface (it will be the radius of the circle whose area is $100m^2$ in your example).

Then you can use trig to solve for $x$, and can then compute the volume of the whole cone (which has height $300+x$ and top radius $y$), and also the smaller cone (which has height $x$ and radius $y$); subtracting these will give you the answer you seek.

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See the explanation in website http://www.lmnoeng.com/Volume/CylConeSphere.htm.

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