5.21 Volumes of RevolutionHL ONLY

1. Revolution around the x-axis

When the region bounded by the curve $y = f(x)$, the x-axis, and the vertical lines $x = a$ and $x = b$ is rotated completely ($360^\circ$ or $2\pi$ radians) around the x-axis, it mathematically generates a 3-dimensional solid.

The volume of this solid of revolution is formulated as:

$$V = \pi \int_{a}^{b} y^2 dx \quad \implies \quad V = \pi \int_{a}^{b} [f(x)]^2 dx$$

Example 1

Find the exact volume of the solid formed when the graph of $y = \sqrt{x}$ from $x=0$ to $x=4$ is rotated $360^\circ$ about the x-axis.

$$ \begin{aligned} V &= \pi \int_{0}^{4} y^2 dx = \pi \int_{0}^{4} (\sqrt{x})^2 dx = \pi \int_{0}^{4} x dx = \pi \left[ \dfrac{x^2}{2} \right]_{0}^{4} \\ &= \pi \left( \dfrac{16}{2} - 0 \right) = \mathbf{8\pi} \end{aligned} $$

2. Volume between two curves

If a region is securely bounded by an upper curve $y = f(x)$ and a lower curve $y = g(x)$, the volume of the solid generated when this region is rotated around the x-axis is evaluated by subtracting the volume of the inner "empty" solid from the outer solid:

$$V = \pi \int_{a}^{b} \left( [f(x)]^2 - [g(x)]^2 \right) dx$$

WARNING: It is a critically common mistake to write $V = \pi \int (f(x) - g(x))^2 dx$. This is mathematically incorrect! You must square the functions individually before subtracting.

Example 2

Find the volume of the solid generated when the region completely bounded by $y = x^2$ and $y = x$ is revolved about the x-axis.

First, we must algebraically find the intersection points to establish the boundaries:
$x^2 = x \implies x^2 - x = 0 \implies x(x - 1) = 0 \iff x = 0 \text{ or } x = 1$
In the interval $[0, 1]$, the upper curve is $y = x$ and the lower curve is $y = x^2$.
$$ \begin{aligned} V &= \pi \int_{0}^{1} \left( y_{\text{upper}}^2 - y_{\text{lower}}^2 \right) dx = \pi \int_{0}^{1} \left( (x)^2 - (x^2)^2 \right) dx \\ &= \pi \int_{0}^{1} (x^2 - x^4) dx = \pi \left[ \dfrac{x^3}{3} - \dfrac{x^5}{5} \right]_{0}^{1} \\ &= \pi \left( \dfrac{1}{3} - \dfrac{1}{5} \right) = \mathbf{\dfrac{2\pi}{15}} \end{aligned} $$

3. Revolution around the y-axis

If the bounded area is instead rotated completely around the y-axis between horizontal boundaries $y = c$ and $y = d$, the integration operates absolutely with respect to $y$:

$$V = \pi \int_{c}^{d} x^2 dy \quad \implies \quad V = \pi \int_{c}^{d} [f(y)]^2 dy$$

Notice that you must mathematically rearrange the structural function to express $x$ exclusively in terms of $y$ before executing the integration.

Example 3

Find the volume of the solid formed when the region bounded by $y = x^3$, the y-axis, and the horizontal lines $y=1$ and $y=8$ is rotated about the y-axis.

We first compute $x^2$. $$ y = x^3 \implies x = y^{1/3} \implies x^2 = \left(y^{1/3}\right)^2 = y^{2/3} $$ Thus, $$ \begin{aligned} V &= \pi \int_{1}^{8} x^2 dy = \pi \int_{1}^{8} y^{2/3} dy = \pi \left[ \dfrac{3}{5} y^{5/3} \right]_{1}^{8} = \pi \left( \dfrac{3}{5}(8)^{5/3} - \dfrac{3}{5}(1)^{5/3} \right) \\ &= \pi \left( \dfrac{96}{5} - \dfrac{3}{5} \right) = \mathbf{\dfrac{93\pi}{5}} \end{aligned} $$