I get the idea of curvilinear coordinates, but don't understand how it would make corresponding symmetrical integrals easier to solve. How would you evaluate an integral, for simplicity's sake say $$ \iint_{S}\textbf{r}\cdot d\textbf{S}, $$ where S is the surface of a sphere with radius R, centered at the origin using polar spherical coordinates?

Would you just convert everything into Cartesians? Wouldn't that just defeat the point of working with curvilinear coordinates?


Well,in this case,you certainly wouldn't. At least,you'd be crazy to do it given how simple the equation becomes in spherical coordinates! This is really a triple integral in $R^{3}$ unless the integral is taken over a level surface.

This is really a poor example-there are many cases where the transformation of coordinates yields a much simpler computation. Several very good examples can be found here. Pay particular attention to the example of the transformation of coordinates for the volume of an ice cream cone. Trying to do this one in Cartesian coordinates would reduce the most battle hardened computational student to tears. The geometry and expression of the final integral in spherical coordinates is amazingly simple. These kinds of transformations arise regularly not only in calculus, but geometry and physics. Indeed-Newton deduced the three laws of motion in classical mechanics from converting the complex equations for the point masses of the planets around the sun into polar coordinates!

That help?

  • $\begingroup$ Yep, thanks a lot! Just out of curiosity, how would you solve the above integral? i.e what would be the uttermost simplest way to calculate it? $\endgroup$ – user167289 Feb 11 '15 at 0:06
  • $\begingroup$ When I try to solve it I get: $\iint_{S}r(\hat{\textbf{r}}\cdot \hat{\textbf{r}})r^2sin\theta d\theta d\phi = \int_{0}^{2\pi}d\phi \int_{0}^{\pi}d\theta r^3sin\theta = 4\pi r^3$ ??? $\endgroup$ – user167289 Feb 11 '15 at 0:12
  • $\begingroup$ That's almost correct,but not quite-you forgot the 1/3 constant factor when you took the antiderivative of r^2 . The volume of a sphere is 4/3 II(r^3). You did it right,you just made a computational error. I wouldn't take off more then a point for that if I was grading your exam.......lol $\endgroup$ – Mathemagician1234 Feb 11 '15 at 2:30

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