SOLUTION: A woodman chops a halfway through a tree having a diameter of 2 meters. One face of the cut being horizontal and the other inclined at 60 degrees. Find the volume of the wood cut

Question 49327: A woodman chops a halfway through a tree having a diameter of 2 meters. One face of the cut being horizontal and the other inclined at 60 degrees. Find the volume of the wood cut out. I know the answer is 1.155 cu. units, but how can we arrive at that answer?
Answer by kev82(151) (Show Source): You can put this solution on YOUR website!
Hi,

(If my coordinate axes seem weird to you, I'm working with x posotive left, y posotive in, z posotive up)

I initially read this and thought it was just some simple volume of revolution problem, however a few diagrams later I realised it was actually quite nasty as there is no rotational symmetry(sp?). I gave up and pulled out one of the the big guns (multiple variable calculus) and got the correct answer of

pretty quickly.

I hope you're at a level where you can follow this, because I have no idea about a simple way to solve it.

First thing is to imagine a circle in the x-y plane radius 1. This is a top down cross-section of our tree trunk. The cut is half way into the tree, so any semicircle will do, but for ease of calculation let's say his horizontal cut is the semicircle given by

.

(I don't know how to draw pictures on here but hopefully you get the idea)

If we now look from the front (the x-z plane) The cross section of the cut is actually a right angled triangle, with the right angle lying on the outer edge of the tree. The angle is of course 60, the adjacent side is x and the opposite side is

This is just basic trigonometry. So if we are at coordinates (x,y,0) the height of the cut above us is

. Now comes the fun part.

Volume as we know is width*depth*height, so if we take lots of cuboids with width dx, depth dy, and height

and integerate them within the limits of our semicircle, we should have the answer. If you've seen multiple inegration before then this is a really easy one and everything should be ok. If not, panic! and then write back.

The integral is clearly

The evaulation is about as straightforward as you can get. The final answer is

. If you've never seen two integrals together like that or need me to explain something in more detail then please write back.

Hope that helps,

Kev

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