# 7) If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.

Error in radius of sphere $(\Delta r)$ = 0.03 m
The surface area of sphere = $4\pi r^2$
Error in surface area $(\Delta A)$
$dA = \frac{dA}{dr}.\Delta r\\ dA = 8\pi r .\Delta r \ \ \ \ \ \ \ \ \ \ \ \ \ (\because A = 4\pi r^2, r =9 \ and \ \Delta r = 0.03 )\\ dA = 8\pi (9) (0.03)\\ dA= 2.16\pi$
Hence,  the approximate error in its surface area is $2.16\pi \ m^2$

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