A cow is tied to a silo with radius r by a rope just long enough to reach the opposite side of the silo. Find the grazing area available for the cow?

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Edit: I found this, would that work in solving the problem?

1 Answer
May 22, 2018

5/6pi^3r^2

Explanation:

First, we need to know the length of rope, and we know the rope can reach the opposite side of the silo.

By letting the length of the rope be l,

l=1/2*2pir
color(white)(l)=pir

https://www.mathalino.com/reviewer/integral-calculus/area-grazing-goat-tied-silohttps://www.mathalino.com/reviewer/integral-calculus/area-grazing-goat-tied-silo

This picture depicts the area the cow can graze and ignore the 10 and replace it with r.

Now to find the area we shall split it into 2 portions,

  1. The semicircle on the left of the diagram

  2. The circle involute on the right of the diagram

1. To find the semicircle,

Find area of the semicircle

A_1=1/2*pi(pir)^2
color(white)(a)=1/2pi^3r^2

https://www.mathalino.com/reviewer/integral-calculus/area-grazing-goat-tied-silohttps://www.mathalino.com/reviewer/integral-calculus/area-grazing-goat-tied-silo

2. To find the circle involute,

Find the length of wrapped rope s,

s=rtheta

Find the length of unwrapped rope r,

r=l-s
color(white)(r)=pir-rtheta
color(white)(r)=r(pi-theta)

Find the area of the circle involute,

A_2=|1/2int_0^pir^2d theta|
color(white)(A_2)=|1/2int_0^pi r^2(pi-theta)^2 d theta|
color(white)(A_2)=|r^2/2[(pi-theta)^3/3]_0^pi|
color(white)(A_2)=|r^2/6[(pi-pi)^3-(pi-0)^3]|
color(white)(A_2)=|-(pi^3r^2)/6|
color(white)(A_2)=(pi^3r^2)/6

https://www.mathalino.com/reviewer/integral-calculus/area-grazing-goat-tied-silohttps://www.mathalino.com/reviewer/integral-calculus/area-grazing-goat-tied-silo

Now, we can find the total area,

A=A_1+2A_2
color(white)(A)=1/2pi^3r^2+2((pi^3r^2)/6)
color(white)(A)=1/2pi^3r^2+1/3pi^3r^2
color(white)(A)=5/6pi^3r^2

Therefore, the grazing area for the cow is 5/6pi^3r^2.