this question is in the book quot;calculus early transcendentalsquot; by james stewart.
because of the fact the cow is ties to a rope that's caught at one end, the cow can be able to pass with the rope in a around way. that's the utmost section that the cow can disguise. component to a circle is pi * r^2 because of the fact the dimensions of the rope is 'r' and that's the gap between the cow and the centre of the circle (the fastened element), the radius of this circle is likewise 'r'. as a result, the section that the cow can disguise is pi * r^2 = 3.14 * r^2
s = radius of the silo. The rope must have a length of πs because it reaches around half the silo. Let θ=0 be the point at which the cow has walked around the silo and is at distance 0 from the silo. If the cow walks around keeping the rope taut then when θ = π the rope is fully extended; if the cow keep walking around the silo, it will eventually be back to 0 extended rope. We can think of this angle is two ways: either sin(θ/2) from 0 to 2π...or...[1 - cos(θ)] which will let us avoid half-angle forms. So, the length of the rope at angle θ is πs[1 - cos(θ)]. Now we are ready to set up our polar coordinate equation and integrate. If r is actually a function of θ, as is the case here, then area = (1/2) I[r^2 dθ]. area = (1/2) I{[πs(1 - cos(θ)]}^2 dθ from 0 to π. We will double our area to account for the full range of motion of the cow. I[1 - cos(θ)]^2 = (3/2)θ - 2sin(θ) + (1/4)sin(2θ), evaulated from 0 to π = 3π/2 --> the area where the cow can walk is (πs)^2*3π/2...*but*...the cow cannot walk inside the silo, so subtract πs^2. Grazing area is thus (πs)^2*3π/2 - πs^2 = πs^2[(3/2)π^2 - 1]. The only unknown is the radius of the silo. update: by the way, the shape of the area where the cow can graze is a cardioid.
