Volume of solid of revolution around axis
Revolve curve around x-axis:
Volume of solid formed =
$ \pi \int_a^b [f(x)]^2 \phantom{.} dx $
Revolve curve around y-axis:
Volume of solid formed =
$ \pi \int_c^d [g(y)]^2 \phantom{.} dy $
Revolve straight lines around axis
Revolving a horizontal/vertical line:
Volume of solid formed =
Volume of cone = $ \pi r^2 h $ = $ \pi (a)^2 (b) $
Revolving a slanted line:
Volume of solid =
Volume of cone = $ {1 \over 3} \pi r^2 h $ = $ {1 \over 3} \pi (a)^2 (a) $
Question
1. The region bounded by the curve $C: y = {1 \over x - 1}$, the normal to the curve $C$ at $x = 2$, the line $x = 3$ and the $x$-axis is rotated through $2 \pi$ radians about the $x$-axis.
Find the volume of the solid generated, leaving your answers in terms of $\pi$.
Answer: $ {5 \over 6} \pi \text{ units}^3$
Solutions
\begin{align*}
\text{Let } x & = 2, \\
y & = {1 \over 2 - 1} \\
y & = 1 \\
\\
\text{Point: } & (2, 1) \\
\\
y & = (x - 1)^{-1} \\
{dy \over dx} & = (-1)(x - 1)^{-2} (1) \\
& = -{1 \over (x - 1)^2} \\
\\
\text{Let } & x = 2, \\
{dy \over dx} & = -{1 \over (2 - 1)^2} \\
{dy \over dx} & = -1 \\
\\
\text{Gradient of normal} & = {-1 \over -1} \\
& = 1 \\
\\
y - y_1 & = m (x - x_1) \\
y - 1 & = (1)(x - 2) \\
y - 1 & = x - 2 \\
y & = x - 1 \phantom{000000} [\text{Equation of normal}]
\end{align*}
\begin{align*}
\text{Volume of solid} & = \underbrace{{1 \over 3} \pi r^2 h}_\text{Cone formed by line} + \underbrace{\pi \int_2^3 \left({1 \over x - 1}\right)^2 \phantom{.} dx}_\text{Solid formed by curve} \\
& = {1 \over 3} \pi (1)^2 (1) + \pi \int_2^3 (1).(x - 1)^{-2} dx \phantom{000000} \left[ \int f'(x) [f(x)]^n \phantom{.} dx = { [f(x)]^{n + 1} \over n + 1} \right] \\
& = {1 \over 3} \pi + \pi \left[ (x - 1)^{-1} \over -1 \right]_2^3 \\
& = {1 \over 3} \pi + \pi [ - (3 - 1)^{-1} ] - \pi [ - (2 - 1)^{-1} ] \\
& = {1 \over 3} \pi - {1 \over 2} \pi + \pi \\
& = {5 \over 6} \pi \text{ units}^3
\end{align*}