wouldn't it be ?If an error Dr is 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 DV with dV, where dV= 4r^2Pidr .
yupReplacing r with 0.5 and dr=Dr with ± 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 ?gives dV=± _______
just choose the larger dV value (larger in terms of absolute value, i would think)The maximum error in the volume is about ________ mm3