Hi all,

After some help with an issue I'm having. Firstly let me say that I haven't learnt any integration of trigonometric functions yet, so I understand I'm doing this a slightly long way. Here is the general synopsis.

I have a cone, which I am trying to find the volume of between two limits. I need to solve this in the general sense, but showing a specific example is easier to explain where I am having the problem. Firstly let me illustrate that I believe the volume of the cone should be 500,000mm^3 within these limits. The cone is not complete, essentially "cut off" below the point with a 20mm radius, like an ice cream cone you've bitten the bottom off.

Volume of maximum cone.
1/3 * \pi * 55.67mm^2 * 161.64mm = 524330.26mm^3

Volume of minimum cone.
1/3 * \pi * 20mm^2 * 58.08mm = 24330.26mm^3

Volume of shape
Maximum cone - minimum cone = 500000mm^3


Now, the function for the line of this cone is y = \tan\theta x with the above cone being the result for 19 degree's.

\tan19 works out to be approximately 0.3443, so as a result the function for this particular line is approximately y = 0.3443x

Now, I wish to integrate this function to get the volume in this cylinder, between the limits of 161.64 and 58.08.

To get the volume of the cylinder from my understanding, we take the integration of f(x)^2 then times it by \pi to get the answer.

As a result we integrate (0.3443x)^2 which gives us 0.11856x^2 to integrate. This results in \frac{0.11856x^3}{3}

Feeding in our limits we get 6813.51mm^2 - 316.16mm^2 = 6497mm^2 under the curve.

We then sub this into \pi * 6497^2 we get the way off answer of 132623725mm^3

What have I done wrong?

Edit: Solved

I was doing two things wrong, firstly I was using the wrong limits in the actual calculations, even though I typed it up correctly here. Secondly I was taking an extra step down the bottom squaring the radius again when I didn't need to, greatly blowing out my answer.

I'll leave it up here even though stupid mistakes are a bit embarrassing, hopefully it might help someone out reading a similar problem to mine.