Volume of Revolution
A solid of revolution is a three-dimensional object obtained by rotating a function in the plane about a line in the plane. The volume of this solid may be calculated by means of integration. Common methods for finding the volume are the disc method, the shell method, and Pappus's centroid theorem.
Volumes of revolution are useful for topics in engineering, medical imaging, and geometry. The manufacturing of machine parts and the creation of MRI images both require understanding of these solids.
Disc Method
Main Article: Disc Method
The disc method calculates the volume of the full solid of revolution by summing the volumes of thin vertical circular disks. This is similar to the notion of integration as being the sum of an infinite number of rectangles. The disc method imagines the solid of revolution as a stack of discs of varying radii. It gives rise to the formula for rotation of the region bounded by \(y = f(x)\), \(y = 0\), \(x = a\), and \(x = b\) about the \(x\)-axis:
\[ V = \int_a^b \pi (f(x))^2 \, dx.\]
Note that the region may be rotated about an arbitrary line \(y = c\) that does not intersect the region with the formula
\[ V = \left| \int_a^b \pi (f(x) - c)^2 \, dx - \int_a^b \pi c^2 \, dx \right|. \]
Shell Method
Main Article: Shell Method
The shell method calculates the volume of the full solid of revolution by summing the volumes of thin cylindrical shells. This gives rise to the formula for rotation of the region bounded by \(y = f(x)\), \(y = 0\), \(x = a\), and \(x = b\) about the \(y\)-axis:
\[ V = \int_a^b 2 \pi x f(x) \, dx.\]
Note that the region may be rotated about an arbitrary line \(x = c\) that does not intersect the region with the formula
\[ V = \int_a^b 2 \pi \lvert x - c \rvert f(x) \, dx. \]
Which Method To Use
When the function \(f(x)\) does not have an inverse with an easily expressible antiderivative, then the disc method should be used for rotations around horizontal lines and the shell method for rotations around vertical lines. In general, the method that gives the simpler integral is preferred.
Determine the volume of the solid obtained by rotating the region bounded by \(y = \frac{1 - x}{1 + x^2}\) and the first quadrant about the \(y\)-axis.
The region in question is bounded by \(y = \frac{1 - x}{1 + x^2}\), \(y = 0\), \(x = 0\), and \(x = 1\). Since \(f(x) = \frac{1 - x}{1 + x^2}\) probably has a somewhat messy inverse, there is no reason not to use the shell method (for rotation about a vertical line, the \(y\)-axis). The volume in question is then
\[\begin{align} \int_0^1 2\pi x \cdot \frac{1 - x}{1 + x^2} \, dx &= 2\pi \int_0^1 \left(\frac{x}{1 + x^2} + \frac{1}{1 + x^2} - 1\right) \, dx \\ \\ &= 2\pi \big(\ln\sqrt{2} + \tfrac{\pi}{4} - 1\big).\ _\square \end{align} \]
Region Bounded By Two Functions
Suppose the region bounded by \(y = f(x)\), \(y = g(x)\), \(x = a\), and \(x = b\) is rotated over a line. Suppose further that \(f(x) \ge g(x)\) for all \(x \in [a, \, b]\).
If the line is vertical \(x = c\) (where \(c \not\in (a, \, b)\)), then the shell method yields a volume
\[ V = \int_a^b 2 \pi \lvert x - c \rvert \big(f(x) - g(x)\big) \, dx. \]
If the line is horizontal \(y = c\) (where \(c \not\in (a, \, b)\)), then the disc method yields a volume
\[ V = \int_a^b \pi \Big\lvert \big(f(x) - c\big)^2 - \big(g(x) - c\big)^2 \Big\rvert \, dx. \]
The volume of the solid formed by revolving the curve
\[y= \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+1}}\]
bounded between \(x = \frac{1}{8}\) and \(x=\frac{9}{16}\) around the \(x\)-axis is in the the form \(A \pi \ln{\frac{B}{C}}.\)
If \(A\) and \(B\) are square-free, what is the value of \(A \times B \times C?\)