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

Explanation

r=9

`deltar`=0.03

we know,

`v=4/3pir^3`

differentiate with respec 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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