# Volumes using Integration

## Volume of Revolution

We can use integration to find volumes of revolution between $x=a$ and $x=b$.

When the region enclosed by $y=f(x)$, the $x$-axis, and the vertical lines $x=a$ and $x=b$ is revolved through $2 \pi$ or $360^{\circ}$about the $x$-axis to generate a solid, the volume of the solid is given by:

$$ \begin{align} \displaystyle

V &= \lim_{h \rightarrow 0} \sum_{x=a}^{x=b}{\pi \big[f(x)\big]^2 h} \\

&= \int_{a}^{b}{\pi \big[f(x)\big]^2}dx \\

&= \pi \int_{a}^{b}{y^2}dx

\end{align} $$

### Example 1

Find the volume of the solid generated when the line $y=x$ for $1 \le x \le 3$ is revolved through $2 \pi$ or $360^{\circ}$ around the $x$-axis.

\( \begin{align} \displaystyle

V &= \pi \int_{1}^{3}{y^2}dx \\

&= \pi \int_{1}^{3}{x^2}dx \\

&= \pi \Big[\dfrac{x^3}{3}\Big]_{1}^{3} \\

&= \dfrac{\pi}{3} \big[x^3\big]_{1}^{3} \\

&= \dfrac{\pi}{3} \big[3^3-1^3\big] \\

&= \dfrac{\pi}{3} \times 26 \\

&= \dfrac{26 \pi}{3} \text{ units}^3

\end{align} \)

### Example 2

Find the volume of the solid generated when the line $y=\sqrt{x}$ for $0 \le x \le 2$ is revolved through $2 \pi$ or $360^{\circ}$ around the $x$-axis.

\( \begin{align} \displaystyle

V &= \pi \int_{0}^{2}{y^2}dx \\

&= \pi \int_{0}^{2}{\sqrt{x}^2}dx \\

&= \pi \int_{0}^{2}{x}dx \\

&= \pi \Big[\dfrac{x^2}{2}\Big]_{0}^{2} \\

&= \dfrac{\pi}{2}\pi \big[x^2\big]_{0}^{2} \\

&= \dfrac{\pi}{2}\pi \big[2^2-0^2\big]_{0}^{2} \\

&= \dfrac{\pi}{2} \times 4 \\

&= 2 \pi \text{ units}^3

\end{align} \)

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