# If the radius of a sphere is measured as $$9 \: m$$ with an error of $$0.03 \: m$$, then find the approximate error in calculating its surface area.

This question has appeared in model paper 2012

Toolbox:
• Let $y=f(x)$
• $\Delta x$ denote a small increment in $x$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
• Surface area=S=$4\pi r^2$
Step 1:
Radius of the sphere =$9m$
Error in measurement =$\Delta r$
$\qquad\qquad\qquad\quad\;\;=0.03m$
Surface area=S=$4\pi r^2$
$S=4\pi r^2$
$\large\frac{dS}{dr}$$=8\pi r[Differentiating with respect to r] Step 2: \Delta S=\large\frac{ds}{dr}$$\times \Delta r$
$\quad\;\;=8\pi r\times 0.03$
$\quad\;\;=8\pi \times 9\times 0.03$
$\quad\;\;=72\pi \times 0.03$
$\quad\;\;=2.16\pi m^2$

edited Aug 6, 2013