Q. 6 - Calculus

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.

Step-by-Step Solution

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 1: Given Information :

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

Step 2: 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 

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