In order to find where to make the cut, we must divide the circle into three very odd shaped pieces. Since the area of the circle is 36pi square inches, each of the thirds must be 12pi square inches.
For simplicity's sake, lets look at quadrant one of the circle. It is one fourth of the circle and spans from (0,6) to (6,0). One of the cuts must be made somewhere on this interval, and the area bounded by the cut and the y axis is one quarter of one of the three pieces you end up with when the cuts are made. This means the area under the curve
y= Sqrt(36-x^2) from 0 to c, (c being the spot of the the cut), must be 3pi. (Since an entire third of the circle is 12pi and this quadrant has one quarter of the center section.)
To find the area under the curve, we must first integrate the equation. To do this, I must consult my integration tables. The integral of an equation of the form Sqrt(a^2 - u^2) when a is a number, (in this case 6) and u is a variable(x) is given as:
.5( x(sqrt(6^2-x^2)) + .5(6^2)(arcsin(x/6)) ) plus a constant. or
.5( x(sqrt(36-x^2)) + 18(arcsin(x/6)) )
To find the area under the curve, we use the definite integral from zero to c. Evaluating the above integral when x = c and subracting the integral when x = 0. The integral when x = 0 is zero, so the area under the curve from 0 to c is expressed as the following.
.5( c(sqrt(36-c^2)) + 18(arcsin(c/6)) )
The area under the curve is equal to 3pi, as stated above. Using a computer or calculator, one can calculate the intersect of the large equation and the line y = 3pi, and solve for c. C is about equal to 1.5896 inches. So the two cuts must be made at 1.5896 and -1.5896 . |
