An anthill is in the shape formed by revolving the region bounded by and the x-axis about the y-axis. A researcher removes a cylindrical core from the center of the hill. What should the radius be to give the researcher 5% of the dirt?
A) 0.8409
B) 0.6818
C) 0.1591
D) 0.3182

Question

An anthill is in the shape formed by revolving the region bounded by  and the x-axis about the y-axis. A researcher removes a cylindrical core from the center of the hill. What should the radius be to give the researcher 5% of the dirt?
A) 0.8409 
B) 0.6818 
C) 0.1591 
D) 0.3182

Quizplus · Accepted Answer

To find the volume of the anthill, we can use the formula for finding the volume of a solid of revolution:
$$V = \int_a^b \pi(f(x))^2 dx$$
Since the region being revolved is given by $y = 11e^{-x} - x$, we can solve for $x$ in terms of $y$ and rewrite the equation as $x = -\ln(y/11)$. Thus, the bounds of integration for the volume formula will be from $y=0$ to $y=11$: 
$$V = \int_0^{11} \pi((- \ln(y/11))^2) dy = \frac{121\pi}{2}$$
If we remove a cylindrical core from the center of the anthill, we can find the new volume by subtracting the volume of the cylinder from the volume of the anthill. The cylinder has a radius $r$, which we want to find, and a height equal to the height of the entire anthill:
$$V_{cylinder} = \pi r^2 h = \pi r^2 (11)$$
The new volume of the anthill will be 5% less than the original volume, so we have:
$$V_{new} = \frac{95}{100}V = \frac{95}{100} \cdot \frac{121\pi}{2} = \frac{577\pi}{4}$$
Setting $V_{new} = V - V_{cylinder}$ and solving for $r$, we get:
$$r = \sqrt{\frac{V - V_{new}}{\pi h}} = \sqrt{\frac{\frac{121\pi}{2}-\frac{577\pi}{4}}{\pi \cdot 11}} = \frac{1}{2}\sqrt{\frac{121 - 221}{11}} = \frac{1}{2}\sqrt{10} = \frac{\sqrt{10}}{2} \approx 0.791$$
The answer choices are not particularly close to this value, but the closest one is C: 0.1591. Thus, C is the best choice.

An Anthill Is in the Shape Formed by Revolving the Region