Question

(c) Use the Mean Value Theorem to show that the volume in part (b) can be written as \(\Delta V = {\tilde \rho ^2}sin\tilde f\Delta \rho \Delta \theta \Delta f\) where \(\tilde \rho \) lies between \({\rho _1}\)and \({\rho _2},\tilde \phi \) lies between \({\phi _1}\)and \({\phi _2},\Delta \rho = {\rho _2} - {\rho _1},\Delta \theta = {\theta _2} - {\theta _1}\) and \(\Delta \phi = {\phi _2} - {\phi _1}\)

Short answer

The result can be written as \({\bar \rho ^2}\sin \bar \phi \Delta \rho \Delta \phi \Delta \theta \)

Step-by-step solution

Define mean value theorem.

Mean Value Theorem stands that for a function, which has derivative at \((a,b)\) and continuous at \((a,b)\) holds

\(\frac{{f(b) - f(a)}}{{b - a}} = {f^\prime }(c)\) where \(a < c < b.\)

Step 2: Now find the volume change.

We have\(\Delta V = \frac{{\rho _2^3 - \rho _1^3}}{3}\left( {\cos {\phi _1} - \cos {\phi _2}} \right)\left( {{\theta _2} - {\theta _1}} \right).\)

By using mean value theorem

\(\begin{aligned}\frac{{\rho _2^3 - \rho _1^3}}{{{\rho _2} - {\rho _1}}} &= 3{{\bar \rho }^2} \Rightarrow \rho _2^3 - \rho _1^3\\ &= 3{{\bar \rho }^2}\Delta \rho \\&\left( {{\rho _1} < \bar \rho < {\rho _2}} \right)\frac{{\cos {\phi _1} - \cos {\phi _2}}}{{{\phi _1} - {\phi _2}}}\\ &= - \sin \bar \phi \Rightarrow \cos {\phi _1} - \cos {\phi _2}\\ &= - \sin \bar \phi \cdot ( - \Delta \phi )\\ &= \sin \bar \phi \Delta \phi \left( {{\phi _1} < \bar \phi < {\phi _2}} \right){\theta _2} - {\theta _1}\\ &= \Delta \theta \end{aligned}\)

Substituting all this in the formula for \(\Delta V\)

we get

\(\begin{aligned}\Delta V &= \frac{1}{3}*3{{\bar \rho }^2}\Delta \rho *\sin \bar \phi \Delta \phi *\Delta \theta \\ &= {{\bar \rho }^2}\sin \bar \phi \Delta \rho \Delta \phi \Delta \theta \end{aligned}\)

Thus, the result can be written as \({\bar \rho ^2}\sin \bar \phi \Delta \rho \Delta \phi \Delta \theta \)