Suppose a cow is tied to a cylindrical silo of radius 2 with a rope which is long enough to go exactly halfway around the silo, so length of rope = 2 pi.what is the total area the cow can reach?
The answer appears to be (10/3) π?. This involved a semicircular region of radius 2π, plus two axe-blade shaped regions, each of area (2/3) π?. Justification follows. Imagine the silo to be a circle of radius 2 centered at (0,2) on the y-axis, so that it passes through the origin O. Let the cow be tethered at O. Then the cow can access a semicircular region in the third and fourth quadrants of radius 2π, i.e. area 2 π?. In the first quadrant, the region accessible to the cow is swept by the rope when it follows the arc of the silo through an angle θ, where 0 ≤ θ ≤ π, and then extends tangentially to a length of 2 (π - θ). As θ increases by an infinitesimal amount dθ, the rope covers a region approximable by an isosceles triangle with two arms of length 2 (π - θ), with an angle dθ between them. The approximation becomes exact as dθ → 0. This region has area ? [2 (π - θ)]? dθ = 2 (π - θ)? dθ. Integrating this as θ varies from 0 to π gives us (2/3) π?, the area accessible to the cow in the first quadrant. This is also the area accessible in the second quadrant. Adding the areas of the three regions, we get 2 π? + 2 (2/3) π? = (10/3) π?.
area is dA = 0.5 r? dΘ , Θ in [- π,π] , and r? = 2? + ( 2π - 2Θ)? r is the hypotenuse of the right triangle with adjacent being 2 and opposite being rope length - arc length of rope on the cylinder...draw the figure..you can certainly do the integration then subtract the area of the circle
Hm... I like this question, but I don't know how to answer it. It's like part of a Mandelbrot Set ( MInus the Silo) for the area the cow can reach.. I'd like to know the answer to this one too.
