1. ## Calculus:Derivatives:Differential (I am stuck on this question help PLEASE)

In a manufacturing process, ball bearings must be made with radius of 0.4 mm, with a maximum error in the radius of ±0.013 mm. Estimate the maximum error in the volume of the ball bearing.
Solution: The formula for the volume of the sphere is________. If an error ∆r is made in measuring the radius of the sphere, the maximum error in the volume is ∆V=__________.
Rather than calculating ∆V, approximate ∆V with dV, where dV=__________.
Replacing r with___ and dr=∆r with ±____ gives dV= ±____.
The maximum error in the volume is about____mm^3.

2. Hello, nch!

Some of this is just arithmetic . . .

In a manufacturing process, ball bearings must be made with $r \,=\,0.4\text{ mm,}$
with a maximum error in the radius of: . $\Delta r = dr = \pm0.013\text{ mm.}$

Estimate the maximum error in the volume of the ball bearing.

Solution: The formula for the volume of the sphere is: . $\boxed{V \:=\:\tfrac{4}{3}\pi r^3}$

If an error $\Delta r$ is made in measuring the radius of the sphere,
the maximum error in the volume is: . $\Delta V \:=\:\boxed{ {\color{white}XXX} }$
The volume of the correct sphere is: . $V_o \:=\:\tfrac{4}{3}\pi (0.4)^3 \:=\:\frac{0.256}{3}\pi\text{ mm}^3$

With an error of $\Delta r = 0.013$, the radius is: . $r \:=\:0.4 + 0.013 \:=\:0.413\text{ mm.}$
And the volume is: . $V_1 \;=\;\tfrac{4}{3}\pi(0.413)^3 \;\approx\;\frac{0.28178}{3}\pi\text{ mm}^3$

The error in volume is: . $\Delta V \;=\;\frac{0.28178}{3}\pi - \frac{0.256}{3}\pi$

. . Hence: . $\Delta V \;=\;0.00859\pi\text{ mm}^3$

Rather than calculating $\Delta V$, approximate $\Delta V$ with $dV$, where: . $dV \:=\:\boxed{4\pi r^2dr}$

Replacing $r$ with $0.4$,and $dr =\Delta R$ with $\pm0.013$ gives: . $dV \:=\:\boxed{{\color{white}XX}}$
$dV \;=\;4\pi(0.4)^2(0.13) \;=\;0.00832\pi\text{ mm}^3$

3. thankz a lot