Question

What is the percentage error in the surface area of a sphere, when the error in measuring its radius is $ \pm 4\% $?

Answer 1

Hint
As we have been given the error in radius, the error in surface area can be calculated as there is a direct relation between error in surface area.
$\Rightarrow S = 4\pi {r^2}$
Where S is the surface area and r is the radius of the sphere.

Complete step by step answer
The error of measurement of the radius of a sphere is $ \pm 4\% $
The surface area of a sphere is $S = 4\pi {r^2}$.
If we take log on both sides, we get,
$\Rightarrow \log S = \log 4\pi + 2\log r$
And after this, we differentiate the above equation with respect to $dr$.
$\Rightarrow \dfrac{1}{S}\dfrac{{dS}}{{dr}} = 0 + \dfrac{2}{r}$
As the derivative of a constant is 0
$\Rightarrow \dfrac{{d(\log 4\pi )}}{{dr}} = 0$
If we rearrange few things, the above equation can be written as,
$\Rightarrow \dfrac{{dS}}{S} = 2\dfrac{{dr}}{r}$
From this equation, we can say that the error in measurement of the surface is twice the error in measurement of the radius.
We have been given the error in measurement of radius, which is $ \pm 4\% $.
If we substitute the value of the error of measurement of the radius in the above equation we will get the error in measurement of surface area.
$\Rightarrow \dfrac{{dS}}{S} = 2\times( \pm 4\% )$
And after this we get the final answer as,
$\Rightarrow \dfrac{{dS}}{S} = \pm 8\% $.

Additional Information
As we have a relation for error in surface area and radius, we also have a relation between error in volume measurement and error in radius measurement.
The volume of a sphere $V = \dfrac{4}{3}\pi {r^2}$.
Now the relation can be written as
$\Rightarrow \dfrac{{dV}}{V} = 3\dfrac{{dr}}{r}$

Note
We have to derive the relation between the measurement error of surface area and radius carefully as if we make a single mistake while deriving it the whole problem will go wrong.