Measuring Metal Loss in a Mechanical Process

A spherical bearing has a radius of \(3{\,{\rm{cm}}}\) when it is new. Use differentials to approximate the volume of the metal lost after the bearing wears down to a radius of \(2.971{\,{\rm{cm}}}\). Compare the approximation to the actual volume lost.

Solution The volume \(V\) of a sphere of radius \(R\) is \(V=\dfrac{4}{3}\pi R^{3}.\) As a machine operates, friction wears away part of the bearing. The exact volume of metal lost equals the change \(\Delta V\) in the volume \(V\) of the sphere, when the change in the radius of the bearing is \(\Delta R=2.971-3={-}0.029{\,{\rm{cm}}}\). Since the change \(\Delta R\) is small, we use the differential \(dV\) to approximate the change \(\Delta V\). Then \[ \begin{eqnarray*} && \Delta V \approx dV \underset{\underset{\color{#0066A7}{\hbox{\(dV=V^{\,\prime} dR\)}}}{\color{#0066A7}{\uparrow }}}{=} 4\pi R^{2}dR \underset{\underset{\color{#0066A7}{\hbox{\(dR=\Delta R\)}}}{\color{#0066A7}{\uparrow}}}{=} (4\pi )(3^{2})({-} 0.029) \approx {-}3.280 \end{eqnarray*} \]

The approximate loss in volume of the bearing is \(3.28{\,{\rm{cm}}}^{3}\).

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The actual loss in volume \(\Delta V\) is \[ \Delta V=V(R+\Delta R)-V(R)=\dfrac{4}{3}\pi \cdot 2.971^{3}-\dfrac{4}{3}\pi \cdot 3^{3}=\dfrac{4}{3}\pi ( -0.7755) =-3.248{\,{\rm{cm}}}^{3} \]

The approximate change in volume is correct to one decimal place.