If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.

Explanation

`r=9`

`Deltar`=0.03

we know,

`v=4/3pir^3`

differentiate with respect to r

`(Deltav)/(Deltar)=delta/(deltar)(4/3pir^3)`

`Deltav=4pir^2 Deltar`

`Deltav= 4 (3.14)(9^2)(0.03)`

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