Say you have  a standing silo with a diameter of 30', surrounded by grass. You attach a 30' tether to the side of the silo and attach a goat to the other end of the tether. What is the total area of grass that the goat could eat? Assume that the goat's reach equals the length of the tether, 30'. The first part is easy, half a circle or 94.2 Ft/2. Once the tether starts wrapping around the silo the radius is constantly decreasing until it becomes zero.
It is not a trick question but a mathematical/scientific one.

I will wait for the answer to appear on YouTube

May I assume that you are assuming that the goat will only walk in one direction?

This is what I'm referring to. Using a coffee can as a silo I anchored one end of the string to the can(at the dot) and tied a sharpie to the other end. As I drew a line around the can the string shortened as it wrapped around the can, resulting in this shape.

Since the problem is symmetric, you can just figure out one side of the area and multiply by 2 (halfway done, see how easy this is?).

For the first 90 degrees of the goat's travel, you have a quarter circle, so the area that the goat can eat is a quarter of (Pi*r^2).

For the last 90 degrees, we will be looking at an arc with a decreasing radius, which is a bit trickier (as in, I'm not entirely sure this will be correct).

At the goat moves around the last 90 degrees of his arc, the length of the tether available will shorten as a function of the lenth of tether that is wrapped around the circumference of the smaller silo, which is a direct relationship (radius of silo * arc in radians), so we end up with the length of the tether being (30 - 15a), where a is the angle travelled, and the goat reaches the end of its tether after travelling 2 radians around the side of the smaller silo (2 radians is about 115 degrees).

So the area that the goat can eat between the silo and the arc of his tether is a calculus problem using polar coordinates.

According to some googling and the Khan academy, using polar coordinates and knowing the radius (free tether length) as a function of the angle, we get that A = 1/2 ](f(r))^2 d@, where f(r) = 30 - 15@

A = 1/2 * ](30-15@)^2 d@  Where  @ goes from 0 to 2.

simplifying, we get ((15)^2)/2  * ](2-@)^2 d@

or 225/2 *  ](4-4@+@^2) d@

= 225/2 * (4@ -2@^2 +(@^3)/3) evaluated from @=0 to 2

Since every term is 0 at @= 0, we get A =

225/2 * (4*2 - 2*2^2  + 8/3) = 225/2 * (8 - 8 + 8/3) = 300

So, half of the area that the goat can eat is:  1/4 * Pi * (30)^2 + 300

= 706.86 + 300 = 1006.86 sq ft.

and the total area that can be eaten by the goat is 2013.72 sq ft**

If the goat's tether didn't have to go around the silo (let's say that it was just a concrete pad), then the area that the goat could have been eaten would have been 2120 sq ft, simply the area of the 30 ft radius circle minus that of the 15 ft radius circle.

Mike

** this is quite likely wrong (OK - it was definitely wrong the first time, and is still quite likely wrong)

Edit -- the question marks showing were supposed to be symbols for integration and for Theta (angle).  Replaced with ] for integral symbol and @ for Theta

Also, screwed up the integration and a bit of basic algebra because I'm an idiot...

Thanks Mike! Not knowing calculus but using the TLAR(that looks about right) method I concur with your solution.
I figured I would get questions like, "What color is the goat?", or "Is it fescue or rye grass?".

Is the goat tethered on the inside of the silo?  Because it won't eat any grass that way.

OP never said if the tether was attached to the top of bottom of the silo...

Yes I over-analyze shit.  Burned my ass many times in college.

Yeah, my immediate mental image was the tether was at the top of the silo.

And then the answer would be no grass, because I've never seen a 30' diameter silo that wasn't taller than 30 feet, and the goat done got hung.