Question

If the radius of a sphere is measured as $9cm$ with an error of $0.03cm$, then find the approximating error in calculating its volume.

• A. $2.46\pi c{m}^3$
• B. $8.62\pi c{m}^3$
• C. $9.72\pi c{m}^3$
• D. $7.46\pi c{m}^3$

Answer 1

Answer: (C)

We have, $r=9cm$ and $\Delta r=0.03cm$
Let $V$ be the volume of the sphere. Then,
$V=\frac{4}{3}\pi r^3$
$\Rightarrow \frac{dV}{dr}=4\pi r^2$
$\Rightarrow {\left(\frac{dV}{dr}\right)}_{r=9}=4\pi \times 9^2=324\pi$
Let $\Delta V$ be the error in $V$ due to error $\Delta r$ in $r$. Then,
$\Delta V=\frac{dV}{dr}\Delta r$
$\Rightarrow \Delta V=324\pi \times 0.03=9.72\pi c{m}^3$
Thus, the approximate error in calculating the volume is $9.72\pi c{m}^3$.