Moo

This should help you out:

Click here for an animated simulation. Link credit: @Julian Poon

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Full Solution:

Notice the behavior of the rope as it makes an arc around the silo. It is fully unwound as it makes a ninety degree angle with the silo in the diagram above. At that point the rope sweeps equal amounts of area throughout an arc measuring $180$ degrees - forming a semi-circle. We shall compute the area $A_1$ of that semi-circle first. Let length of rope be $r$ for that sector. Then,

$A_1=\frac{\pi r^2}{2}=\frac{\pi L_{\text{rope}}^2}{2} = \frac{\pi (10 \pi)^2}{2}=\boxed{50\pi^3}$

For the other section, we will compute the area by considering the variable radius as the rope swings an arc around the silo. For a specific angle swept by the rope, the length of the rope still wrapped around the silo is

$s=10\theta$ $^*$

Accordingly, the length of the unwrapped portion of the rope will be the radius:

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

For the area $A_2$ swept by the rope, we will take an integral of infinitesimal area increments $dA_2$ in terms of the variable radius $r$ for all the infinitesimal angle increments $d\theta$ from $0$ to $\pi$ . We will then double that area to obtain the full right-hand sector:

$A_2=2\left(\frac{1}{2}\int_{\theta_1}^{\theta_2}{r^2} d\theta \right) ^{**}$

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

$A_2=100\left[ \frac{(\pi - \theta)^3}{3} \right]^{\pi/2}_0$

$A_2=\boxed{\frac{100}{3}\pi^3}$

Thus, the total area for grazing is

$A=A_1+A_2=50\pi^3+\frac{100\pi^3}{3}=\frac{250}{3}\pi^3=\boxed{2583.86}$

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$*$ - This is from the fact that $C=2\pi r$ . Here, $2\pi$ represents $360$ degrees in radians - a full circle. We replace that with $\theta$ so that we compute the length of the rope still wrapped around the silo for any angle swept. The general formula is $s=r\theta$ - the expression for arc length.

$**$ - This integral is actually a double integral:

$2\left(\int_{\theta_1}^{\theta_2}{\left( \int_{0}^{r}{\left( r \right )} dr \right) } d\theta \right)$

Here we are integrating circumference with respect to radius. We do this by summing infinitesimal area increments $dA$ produced by the product of the infinitesimal radius increments $dr$ and the circumference - from $0$ to $r$ , like this:

(Here $dr$ and $dA$ are exaggerated for purposes of visualization).

Notice if we had $2\pi r$ in the integral, we'd produce $\pi r^2$ - which is the area of a full circle. Except here we don't have a full circle. We don't use the constant value for circumference - $2\pi$ . We are concerned with specific variable radii for their respective sweeps $d\theta$ , which is why we don't include the $2\pi$ in the integral.

Also notice that this does not produce a volume, for the radial displacement $d\theta$ is dimensionless. The double integral serves to account for the two varying quantities - the radius $r$ and the angle swept $\theta$ . This way we're summing the product of the specific length of the rope $r$ as it sweeps through an angle $d\theta$ - producing an area $dA$ . The sum of all those areas produces the full area swept by the rope for that sector.

$Area ~covered =\dfrac 5 6*\dfrac {L^3*\pi} r.~~ Our~~ r=10,~ and~ L=\pi*r.~~ So~ the~ area=\dfrac 5 6*pi^3*r^2=\dfrac 5 6 *pi^3*10^2=2583.86.$

(pi*(10pi)^2)/2 + integral [0 to 10pi] (10pi-X)dX = 50pi^3 + (10piX-(X^2/2))[0 to 10pi] = 50pi^3 + (100pi^2-(100pi^2)/2) = 2537.27