ok

wouldn't it be ?If an error Dris made in measuring the radius of the sphere, the maximum error in the volume is DV=4(r+delta(r))^2pi-4^2pi

okRather than calculating DV, approximate DVwithdV, wheredV=4r^2Pidr.

yupReplacingrwith0.5anddr=Drwith ±0.019

just plug in the values and calculate this. you have the formula for dV, what do you get when you plug in r = 0.5 and dr = 0.019, and then when you plug in r = 0.5 and dr = -0.019 ?givesdV=± _______

just choose the larger dV value (larger in terms of absolute value, i would think)The maximum error in the volume is about ________ mm3