Re: Find Dimensions of a Log

Hey BobBali.

The thing where you slipped up was when you equated the area of the circle with the area of the square: you need to account for the left over area.

I am going to use a technique involving integration to get the value of the square length.

Let x^2 + y^2 = r^2 denote the relationship of a circle at the origin with radius r. We will now solve a boundary value problem where

2*Integral (-r,r) SQRT(y^2 - x^2)dx = Area_Of_Circle = x^2 + LeftOverArea = 400*pi.

The "LeftOverArea" is the area that is not part of the square, but part of the circle.

We have 2*[Integral(-r,-r+y0)SQRT(r^2 - x^2)dx + Integral(r-y0,r)SQRT(r^2 - x^2)dx] = LeftOverArea = 400*pi - x^2

By solving LeftOverArea, we solve x, and do to this, we solve for y0 where x = r - y0.

I will complete this a little later on, but if you can pick up where I am going with this then feel free to complete it yourself.

Re: Find Dimensions of a Log

using simple geometry I get an answer of 28.3 cm