Cowculus

the area can be divided into 3 parts. let the radius of the circular barn be r unit, the length of rope will be $\pi r$ unit.

$A_3 = \frac{1}{2} \pi (\pi r)^2 = \frac{1}{2} \pi^3 r^2$ .

for $A_1$ and $A_2$ , $\begin{aligned} dS &= \frac{1}{2} (\pi r - r \theta)^2 \ d\theta \\ &= \frac{1}{2} r^2 (\pi - \theta)^2 \ d\theta \\ \int_0^{A_1} dS &= \frac{1}{2} r^2 \int_0^{\pi} (\pi - \theta)^2 \ d\theta \\ A_1 &= \frac{1}{6} \pi^3 r^2 \\ A_2 &= A_1 = \frac{1}{6} \pi^3 r^2 \\ \therefore \text{area} &= \frac{1}{2} \pi^3 r^2 + \frac{1}{6} \pi^3 r^2 + \frac{1}{6} \pi^3 r^2 \\ &= \frac{5}{6} \pi^3 r^2 \\ \text{subs } r=6, \text{ area} &=30\pi^3 \\ \therefore A &= 30 \end{aligned}$

Here's my approach in brief:

Image made in GeoGebra graphing calculator

• find parametric equation of the red curve: $\begin{aligned} x(t) &= R(\sin{t} + (\pi - t)\cos{t}) \\ y(t) &= R(1-\cos{t} + (\pi - t)\sin{t})\end{aligned}$

• calculate the green area as: ${\color{#456461} \text{Area}} = \int_{\pi}^{0}y(t)\cdot x'(t)\,dt - \dfrac{R^{2}\pi}{2} = \dfrac{R^{2}\pi^{3}}{6}$ I used WolframAlpha to evaluate this definite integral.

• calculate the whole area as: $2\times{\color{#456461} \text{Area}} + \dfrac{(R\pi)^2\pi}{2} = \dfrac{5}{6}R^{2}\pi^{3}.$ From here, we deduce that $A = 30$ .