Recall that the area bounded by a curve, the x-axis, and the two lines $x=a$ and $x=b$, is given by:
$$ \begin{equation}\begin{aligned} A=\int_a^bf(x)dx\\ \end{aligned}\end{equation} $$If we were to take this area and rotate $360\degree$, the volume of the solid shape generated by this rotation would be given by:
$$ \begin{equation}\begin{aligned} V=\pi\int_a^b[f(x)]^2dx\\ \end{aligned}\end{equation} $$Note the two changes:
- We multiply by pi ($\pi$)
- We integrate the square of the function instead of just the function
Example
FInd the volume of revolution when the area bounded by the curve $y=x+3$, and the lines $x=1$ and $x=3$, and the x-axis, is rotated about the x-axis.
Solution The volume is given by the integral: $$ \begin{equation}\begin{aligned} V&=\pi\int_a^b[f(x)]^2dx\\ &=\pi\int_1^3[(x+3)]^2dx\\ &=\pi\int_1^3(x+3)^2dx\\ &=\pi \left[ \frac{(x+3)^3}{3\cdot 1} \right]_{1}^3\\ &=\pi \left[ \frac{(x+3)^3}{3} \right]_{1}^3\\ &=\pi \left[\frac{(3+3)^3}{3}-\frac{(1+3)^3}{3}\right]\\ &=\pi\left[\frac{216}{3}-\frac{64}{3}\right]\\ &=\frac{152\pi}{3} \text{units}^3\\ \end{aligned}\end{equation} $$
Volume about the y-axis
The same principles can be applied to a rotation about the y-axis but we must have everything in terms of $y$ instead - a function written in terms of $y$, two $y$ boundaries ($y=a$ and $y=b$) and a $dy$ to signify the y-axis: $$ \begin{equation}\begin{aligned} V=\pi\int_a^b[f(y)]^2dy\\ \end{aligned}\end{equation} $$