# Area for grazing by the goat tied to a silo

**Problem**

A goat is tied outside a silo of radius 10 m by a rope just long enough for the goat to reach the opposite side of the silo. Find the area available for grazing by the goat. Note that the goat may not enter the silo.

A. 1033.54 m^{2}

B. 2067.08 m^{2}

C. 516.77 m^{2}

D. 2583.86 m^{2}

**Solution**

$L_{rope} = 10\pi ~ \text{m}$

The total area for grazing is shown above. It is composed of a semi circle and area bounded by involute of circle, the circle, and the vertical line along the quadrant point on the circle.

**For the semi-circle**

$A_1 = \frac{1}{2}\pi (L_{rope})^2 = \frac{1}{2}\pi (10\pi)^2 $

$A_1 = 50\pi^3 ~ \text{m}^2$

**For the involute of circle**

Length of rope wrapped on the silo

$s = 10\theta$

Length of unwrapped/stretched rope

$r = L_{rope} - s = 10\pi - 10\theta$

$r = 10(\pi - \theta)$

$\displaystyle A_2 = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta$

$\displaystyle A_2 = \frac{1}{2} \int_{0}^{\pi} 100(\pi - \theta)^2 \, d\theta$

$\displaystyle A_2 = 50 \int_{0}^{\pi} (\pi - \theta)^2 \, d\theta$

$\displaystyle A_2 = -50 \int_{0}^{\pi} (\pi - \theta)^2 \, (-d\theta)$

$A_2 = -50 \left[ \dfrac{(\pi - \theta)^3}{3} \right]_{0}^{\pi/2}$

$A_2 = -\frac{50}{3} \left[ (\pi - \pi)^3 - (\pi - 0)^3 \right]$

$A_2 = -\frac{50}{3} (-\pi^3)$

$A_2 = \frac{50}{3}\pi^3 ~ \text{m}^2$

**Total area for grazing**

$A = A_1 + 2A_2$

$A = 50\pi^3 + 2(\frac{50}{3}\pi^3)$

$A = \frac{250}{3}\pi^3 = 2\,583.86 ~ \text{m}^2$ *answer: D*

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