Question

If the radius of the sphere is measured as 9 m with an error of 0.03 m , then find the approximate error in calculating the surface area?

Answer 1

We know that the surface area of the sphere is $4\pi {r^2}$ and the error in this surface area is given by differentiating the surface area with respect to r and substituting the given values to obtain the approximate error.

Complete step-by-step answer:
We are given the radius and the error in measuring the radius.
Let r be the radius and $\vartriangle r$ be the error in measuring the radius
Then , r = 9 m and $\vartriangle r = 0.03m$
We know that the surface area of the sphere is given by $4\pi {r^2}$
$\Rightarrow S = 4\pi {r^2}$
Now lets differentiate the surface area with respect to r
$\Rightarrow \dfrac{{dS}}{{dr}} = \dfrac{{d(4\pi {r^2})}}{{dr}} = 8\pi r$
Now the error in surface area is given by
$\Rightarrow dS = 8\pi r.\vartriangle r$
Now substituting the given values we get
$\Rightarrow dS = 8\pi {(9)^2}{(0.03)^2} \\\ \Rightarrow dS = 2.16\pi {m^2} \\\$
Hence the approximate error in calculating the Surface area is $2.16\pi {m^2}$.

Note: The approximation error in some data is the discrepancy between an exact value and some approximation to it. An approximation error can occur because the measurement of the data is not precise due to the instruments. (e.g., the accurate reading of a piece of paper is 4.5 cm but since the ruler does not use decimals, you round it to 5 cm.) or approximations are used instead of the real data