A solid is generated by revolving the region bounded by $y = \frac { 1 } { 12 } x ^ { 2 } \text { and } y = 12$ about the y-axis. A hole, centered along the axis of revolution, is drilled through this solid so that one-fourth of the volume is removed. Find the diameter of the hole. Round your answer
To three decimal places.
A)10.733
B)11.167
C)4.392
D)2.536
E)8.785

Question

Quizplus · Accepted Answer

First, we need to find the total volume of the solid. We can do this using the formula for volume of a solid of revolution:

$$V = \int_{a}^{b} \pi (f(x))^2 dx$$

where $a$ and $b$ are the bounds of integration (in this case, $0$ and $24$ since the graphs intersect at $(24, 12)$), and $f(x)$ is the function being revolved. In this case, we have two functions, so we need to split the integral into two parts:

$$V = \int_{0}^{24} \pi \left(\frac{1}{12}x^2ight)^2 dx - \int_{0}^{24} \pi 12^2 dx$$

Evaluating each integral, we get:

$$V = \frac{1}{144} \pi \int_{0}^{24} x^4 dx - 12^2 \pi \int_{0}^{24} dx$$

$$V = \frac{1}{144} \pi \left[\frac{1}{5}x^5ight]_{0}^{24} - 12^2 \pi \left[xight]_{0}^{24}$$

$$V = \frac{1}{144} \pi \cdot 796,262.4 - 12^2 \pi \cdot 24$$

$$V = 122,472.056 	ext{ cubic units}$$

One-fourth of this volume is removed, so the new volume is:

$$V_{	ext{new}} = \frac{3}{4} V$$

$$V_{	ext{new}} = 91,854.042 	ext{ cubic units}$$

Now, we need to find the radius of the hole. Let the radius be $r$, and let $h$ be the height of the cylindrical hole (which we don't know yet).

The volume of the hole can be found using the formula for volume of a cylinder:

$$V_{	ext{hole}} = \pi r^2 h$$

The volume of the solid with the hole can be found using the same formula as before, but with the smaller volume limit:

$$V_{	ext{new}} = \int_{0}^{24} \pi (f(x))^2 dx - V_{	ext{hole}}$$

Substituting in the functions and evaluating the integral, we get:

$$V_{	ext{new}} = \frac{1}{144} \pi \cdot 796,262.4 - \pi r^2 h$$

Setting this equal to the volume of the solid with one-fourth of the volume removed, we can solve for $r$:

$$\pi r^2 h = \frac{1}{4} \left(\frac{1}{144} \pi \cdot 796,262.4ight)$$

$$\pi r^2 h = 22,993.511 	ext{ cubic units}$$

$$r^2 h = 7,339.88$$

Substituting in the equation for $f(x)$, we can solve for $h$ in terms of $r$:

$$h = 12 - \frac{1}{12}(\sqrt{12^2 - r^2})^2$$

$$h = 12 - \frac{1}{12}(144 - r^2)$$

$$h = \frac{1}{12}(r^2)$$

Substituting this into the equation for $r^2h$, we get:

$$r^4 = 7,339.88 \cdot 12$$

$$r^4 = 88,078.56$$

$$r = \sqrt[4]{88,078.56} \approx 8.785$$

Therefore, the diameter of the hole is twice the radius, or approximately $\boxed{	extbf{(E) } 8.785}$.

A Solid Is Generated by Revolving the Region Bounded By y