# Volumes for Two Functions

If the region bounded by the upper function $y_{upper}=f(x)$ and the lower funciton $y_{lower}=g(x)$, and the lines $x=a$ and $x=b$ is revolved about the $x$-axis, then its volume of revolution is given by:
\begin{align} \displaystyle V &= \int_{a}^{b}{\Big([f(x)]^2-[g(x)]^2\Big)}dx \\ &= \int_{a}^{b}{\Big(y_{upper}^2-y_{lower}^2\Big)}dx \end{align}

### Example 1

Find the volume of revolution generated by revolving the region between $y=x^2$ and $y=\sqrt{x}$ about the $x$-axis.

\begin{align} \displaystyle x^2 &= \sqrt{x} \\ x^4 &= x \\ x^4-x &= 0 \\ x(x^3-1) &= 0 \\ x &= 0 \text{ and } x=1 \\ V &= \pi \int_{0}^{1}{\big[\sqrt{x}^2-(x^2)^2\big]}dx \\ &= \pi \int_{0}^{1}{(x-x^4)}dx \\ &= \pi \bigg[\dfrac{x^2}{2}-\dfrac{x^5}{5}\bigg]_{0}^{1} \\ &= \pi \bigg[\dfrac{1^2}{2}-\dfrac{1^5}{5}\bigg]-\pi \bigg[\dfrac{0^2}{2}-\dfrac{0^5}{5}\bigg] \\ &= \dfrac{3 \pi}{10} \text{ units}^3 \end{align} ## Trigonometry Made Easy: Integration by Parts Demystified

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