Surface area of a solid of revolution obtained by revolving the graph of a nonegative differentiable function about the x-axis Choose a number $x$ in the interval $[a,b]$ and let $A(x)$ be the area of the part of the solid of revolution to the left of the vertical plane through $(x,0,0)$. Then $A(a)=0$ and $A(b)=\textrm{surface area of the solid of revolution }L$. For a small increment $\dd x$ the increment $A(x+\dd x)-A(x)$ is equal to the area of the thin band of the solid of revolution between the vertical planes through the points $(x,0,0)$ and $(x+\dd x,0,0)$. See the figure below.

Since the left edge of the band is a circle wih radius $f(x)$, its circumference is $2\pi f(x)$. So we still need to determine the width $\dd s$ of the band to compute its surface area. For very small (mathematicians say 'infinitesimally small') increment $\dd x$ we almost have a straight line segment $\dd s$ and the area of the band is equal to $2\pi f(x)\cdot \dd s$. The line segment $\dd s$ has length

$$\sqrt{\dd x^2+\dd f^2}=\sqrt{\dd x^2+\bigl(f'(x)\,\dd x\bigr)^2}=\sqrt{1+\bigl(f'(x)\bigr)^2}\dd x$$ The expression $2\pi f(x)\sqrt{1+\bigl(f'(x)\bigr)^2}\dd x$ is a differential and equal to the differential $\dd A$ of the area function. So: $\dd A=2\pi f(x)\sqrt{1+\bigl(f'(x)\bigr)^2}\dd x$.

The requested surface area of the solid of revolution $S$, which is obtained by revolving the graph of the function $f$ on the interval $[a,b]$ about the $x$-axis, is equal to

$$\begin{aligned}\text{surface area of }S&=A(b)\\[0.25cm]&=\int_a^b 2\pi f(x)\,\sqrt{1+\bigl(f'(x)\bigr)^2}\dd x\\[0.25cm]&=2\pi\int_a^bf(x)\,\sqrt{1+\bigl(f'(x)\bigr)^2}\dd x\end{aligned}$$

Surface area of a field of revolution obtained by revolving the graph of a nonegative differentiable function about the y-axis The surface area of the solid of revolution $S$ obtained by revolving the graph of a nonegative differentiable function $f$ on the interval $[a,b]$ about the $y$-axis is equal to

$$\begin{aligned}\text{surface area of }S=2\pi\int_a^b x\,\sqrt{1+\bigl(f'(x)\bigr)^2}\dd x\end{aligned}$$

Surface area of a solid of revolution obtained by revolving a parametric curve about the $x$- or $y$-axis When a parametic curve $\bigl(x(t),y(t)\bigr)$ on the interval $[a,b]$ is revolved about the $x$- or $y$-axis, then the surface area $A(S)$ of the solid revolution $S$ can be computed by the formulas below.