1.

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

Answer»

Let r be the radius of the sphere and ∆r be the error in measuring the radius then r =7m and ∆r = 0.02 m

We have;

V = \(\frac{4}{3}\) πr3
dV = \(\frac{dV}{dr}\) ∆r = \(\frac{4}{3}\) π3r2 × ∆r
= 4π(7)2 × 0.02 = 3.92 π m3.



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