Question

If the radius of a sphere is measure as 7m with an error of 0.02 m, then find the approximate error in calculating its volume.

Answer 1

Let r be the radius of the sphere and $\Delta r$ be the error in measuring radius.
Then, r = 7 m and $\Delta r=0.02m.$
Now, volume of a sphere is given by $V=\frac{4}{3}\pi r^3$
On differentiate w.r.t.r, we get $\frac{dV}{dr}=\left(\frac{4}{3}\pi \right)\left(3r^2\right)=4\pi r^2$
$\therefore \Delta V=\left(\frac{dV}{dr}\right)\Delta r=\left(4\pi r^2\right)\Delta r=4\pi \times 7^2\times 0.02=3.92\pi m^3$
Hence, the approximate error in calculating the volume is $3.92\pi m^3$