Question

Write down definite integrals to express the given geometric quantities:

The area of the surface is obtained by revolving $f\left(x\right)$on $\left[a,b\right]$around the x-axis.

Short answer

Surface area of$f\left(x\right)$on$\left[a,b\right]$$=2\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\sqrt{1+{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2}d\mathrm{x}$

Step-by-step solution

Given Information :

Given that to assume that $f\left(x\right)$is continuous and differentiable, with a continuous derivative.

Definite integral formula for surface area :

The surface area S of the solid of revolution obtained by revolving $y=f\left(x\right)$around the x-axis from $x=a$ to $x=b$ is

$S=2\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\sqrt{1+{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2}d\mathrm{x}$