Problem 48 - Calculus of a Single Variable

Question

Machine Part A solid is generated by revolving the region bounded by \(y=\sqrt{9-x^{2}}\) and \(y=0\) about the \(y\) -axis. A hole, centered along the axis of revolution, is drilled through this solid so that one-third of the volume is removed. Find the diameter of the hole.

Step-by-Step Solution

Answer

The diameter of the hole drilled in the solid is approximately 3.36 circumference units.

Step 1: Determine the total volume of the solid

To calculate the volume, we first need to consider the function \(y=\sqrt{9-x^{2}}\) which describes a semicircle of radius 3. When this region is revolved about the y-axis, it forms a solid hemisphere of radius 3. We know that the volume \(V\) of a sphere of radius \(r\) is given by \(\frac{4}{3}\pi r^{3}\). So, the volume of a hemisphere would be half of that. In this case, \(V_{total} = \frac{1}{2}(4/3)\pi(3)^{3} = \frac{18}{2}\pi = 9\pi\).

Step 2: Determine proportion of volume removed and interpret it as a hemispherical hole

It is specified that a third of the total volume is removed by drilling a hole through the solid. This corresponds to a volume \(V_{hole}= \frac{1}{3} * V_{total}\). This volume can be interpreted as a solid hemisphere of an unknown radius \(r_{hole}\) drilled into the solid, thus: \(V_{hole} = \frac{1}{2}(4/3)\pi r_{hole}^{3}\).

Step 3: Find the radius of the drilled hole

Set the equalities from Step 2: \( \frac{1}{3} * V_{total} = \frac{1}{2}(4/3)\pi r_{hole}^{3}\). Substituting \(V_{total} = 9\pi\), we get, \(3\pi = \frac{2}{3}\pi r_{hole}^{3}\). Simplifying, we find \(r_{hole}^{3}= \frac{9}{2}\). Taking the cube root from both sides, we get \(r_{hole} = \sqrt[3]{\frac{9}{2}}\) or \(1.68\) (approximately).

Step 4: Determine the diameter of the hole

Since diameter is twice the radius, the diameter of the drilled hole \(d_{hole}=2*r_{hole} = 2*\sqrt[3]{\frac{9}{2}}\) or approximately \(3.36\) circumference units.