I Calculating the change of the volume of a sphere using this integral

[cppIStough]

I have a spherical cap of liquid (drop) that rests on a substrate. The substrate has a small hole at the base where liquid is pumped into the drop. One way to parameterize the spherical cap is via: $$x = \sin(s)\cos\phi/\sin\alpha, \,\,\,y= \sin(s)\sin\phi/\sin\alpha,\,\,\,z=(\cos(s)-\cos\alpha)/\sin\alpha$$ where here $\alpha$ is the angle the spherical cap makes with the x-y plane, and $\phi \in [0,2\pi]$ and $s\in[0,\alpha]$.

Now $\eta$ is a small disturbance to the drop. The author then states the volume change can be written as $$\Delta V = \int_0^{2\pi}\int_{\cos\alpha}^1 R^2 \eta d(\cos(s))d\phi$$

Can someone explain the math behind that integral? I understand the $\phi$ integral, but the inside integral I can't make sense of. For help, here's a sketch:

 

[pasmith]

This is a straightforward use of spherical polar coordinates (r, s, \phi):
\begin{split} <br /> \Delta V &amp;= \int_0^{2\pi} \int_0^{\alpha} \int_R^{R + \eta} r^2\sin s\,dr\,ds\,d\phi \\<br /> &amp;= \int_0^{2\pi} \int_0^\alpha \frac13\left( (R + \eta)^3 - R^3 \right) \sin s\,ds\,d\phi \\<br /> &amp;= \int_0^{2\pi} \int_{\cos \alpha}^1 \frac13\left( 3R^2\eta + 3R\eta^2 + \eta^3\right)\,d(\cos s)\,d\phi.<br /> \end{split} Since \eta is a small perturbation, only the term \frac13 (3R^2\eta) is retained.

 

[cppIStough]

This is a straightforward use of spherical polar coordinates (r, s, \phi):
\begin{split}<br /> \Delta V &amp;= \int_0^{2\pi} \int_0^{\alpha} \int_R^{R + \eta} r^2\sin s\,dr\,ds\,d\phi \\<br /> &amp;= \int_0^{2\pi} \int_0^\alpha \frac13\left( (R + \eta)^3 - R^3 \right) \sin s\,ds\,d\phi \\<br /> &amp;= \int_0^{2\pi} \int_{\cos \alpha}^1 \frac13\left( 3R^2\eta + 3R\eta^2 + \eta^3\right)\,d(\cos s)\,d\phi.<br /> \end{split} Since \eta is a small perturbation, only the term \frac13 (3R^2\eta) is retained.

This is great! What made you think to integrate from $R$ to $R + \eta$? Becasue $\eta$ is deviation from equilibrium, so for me it wasn't obvious that it implies a change in volume.

EDIT: except now I look at the form and it's a function of $t$ which of course means it grows. Thanks so much!

 

[pasmith]

If the position of the surface of the drop changes, then the volume enclosed by it will also change.