# If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere.

Given the error in the measurement of the radius of a sphere is 0.1%.

Let x be the radius of the sphere and Δx be the error in the value of x.

Hence, we have Δx = 0.001x

The volume of a sphere of radius x is given by On differentiating V with respect to x, we get  We know   Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as Here, and Δx = 0.001x

ΔV = (4πx2)(0.001x)

ΔV = 0.004πx3

The percentage error is,  Error = 0.003 × 100%

Error = 0.3%

Thus, the error in calculating the volume of the sphere is 0.3%.

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