The radius of a sphere is claimed to be 3.0 inches, correct to within 0.01 inch. use linear approximation to estimate the resulting error, measured in cubic inches, in the volume of the sphere.

0,36π inches³ or simply 1.13 inches³

Step-by-step explanation:

In this case, first, we need to calculate the volume of the sphere with the formula:

V = 4πr³/3

Replacing the data we have:

V = 4π(3)³/3

V = 36 in³

Now that we have the volume we need to calculate the resulting error. In this case, using linear approximation we have to use the derivates of V and r, so we have the following:

If V = 4πr³/3

Then the derivate of V (dV) would be:

dV = 4π*3*r²/3 dr

Where dr is the error of radius so:

dV = 4π*r² dr

Solving for dV:

dV = 4*3.14*(3)²*(0.01)

dV = 1.13 in³

So at the end, we just report the volume of the sphere as

V = 36 ± 1 in³

dV =

i dont know sorry

step-by-step explanation:

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step-by-step explanation: