Question

If the radius of a sphere is measured as $8cm$ with an error of $0.03cm$, then the approximate error in calculating its volume is

• A. $7.62\pi c{m}^3$
• B. $7.68\pi c{m}^3$
• C. $7.86\pi c{m}^3$
• D. $6.68\pi c{m}^3$

Answer 1

Answer: (B)

Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius.
Then $r=8cm$, and $\Delta r=0.03cm$,
Now the volume $V$ of the sphere is given by
$V=\frac{4}{3}\pi r^3$
$\frac{dV}{dr}=4\pi r^2$
$\Delta V=\left(\frac{dV}{dr}\right)\Delta r=\left(4\pi r^2\right)\Delta r$
$=4\pi (8{)}^2\times 0.03=7.68\pi c{m}^3$