If $f(x)\ge 0$ on $[a,b]$ and $f'(x)$ is [[Continuous]] on $[a, b]$, the surface obtained by rotating the curve $y=f(x)$ about the $x$ [[Axis]] from $x=a$ to $x=b$ is: $\huge \text{Surface Area} = \int_{a}^b 2\pi f(x)\sqrt{1+f'(x)^2}\d x $ > [!note]- > This formula extends / is similar to [[Arclength|the formula for an Arclength of a surface of revolution]]. >[!example] Find the area of the surface obtained by rotating the [[Curve]] $y=\sqrt{1+4x}$ about the $x$ [[Axis]] from $x=1$ to $x=5$ >>[!check]- Solution >>$\huge \begin{align} >>f(x) &= \sqrt{1+4x} = (1+4x)^{ \frac{1}{2} } \\ >>f'(x) &= 2(1+4x)^{ -\frac{1}{2} }\\ >>&= \frac{2}{\sqrt{1+4x}}\\ \\ S &= \int_{1}^5 2\pi f(x)\sqrt{1+f'(x)^2}\d x \\ &= \int_{1}^5 2\pi \sqrt{1+4x} \sqrt{1+\frac{4}{1+4x}}\d x \\ &= 2\pi\int_{1}^5 \sqrt{1+4x} \sqrt{\frac{5+4x}{1+4x}}\d x \\ &= 2\pi \int_{a}^5 \sqrt{5+4x} \d x \\ \\ \let u &=5+4x \\ \let \d u &= 4\d x \\ \\ S &= \frac{\pi}{2} \int_{9}^{25}\sqrt{u} \d u \\ &= \frac{\pi}{2} \cdot \frac{2}{3} u^{ \frac{1}{2} } \big \vert^{25}_{9} \\ &= \frac{\pi}{3} \pa{125-27} \\ &= \boxed{ \frac{98\pi}{3} \text{units}^2 } \end{align}$