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**Ash_underpar** Consider a function $\displaystyle f(x)$ where $\displaystyle f(x) \succeq 0$ for $\displaystyle a \preceq x \preceq b$.

Define the 'surface of revolution' to be the surface swept out by a complete revolution of the curve $\displaystyle f(x)$, $\displaystyle a \preceq x \preceq b$, about the x-axis.

This surface of revolution can be described using the parameterization $\displaystyle r(u,v)=[v, f(v)cos u, f(v)sin u]$, $\displaystyle a\preceq v \preceq b,$ $\displaystyle 0 \preceq u \preceq 2\pi$.

Show that the area of this surface is given by $\displaystyle A=2\pi \int_a^b f(v) \sqrt{1+(f'(v))^2}dv$ where $\displaystyle f'(v)=\frac{df}{dv}$