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.