Problems with conversions to spherical coordinates involving a line integral

[akalisman]

Homework Statement

given the vector A = 4r + 3theta -2phi
, find its line integral around the closed path.
(the figure contained in the book is a straight line along the x-axis extending to radius a, with a curved portion of a circle with radius a centered at the origin curving back to the y-axis at y= +a and then another straight line returning to the origin, the direction of the path flows counter-clockwise from the origin). Also find the surface integral of del(cross)A over enclosed area and compare results.

Homework Equations

A = 4r + 3theta -2phi

The Attempt at a Solution

my question regards the setup of the integral, because the A vector is given in spherical coordinates i assume the rest of the integral must also be in spherical, however what would be the ds portion used for the line integral and because the figure is given in cartesian coordinates how do you convert those into spherical coordinates? If i remember correctly it would be split into a summation of 3 different integrals for each separate path, the first and second path should be easy enough given that they are straight lines, but the second path is an arc for which the equation in rectangular coordinates would be a² = x² + y², but how would you convert this to spherical coordinates as well?

 

[HallsofIvy]

On the straight line from (0,0,0) to (a, 0, 0), \rho goes from 0 to a, \theta= 0 and \phi= \pi. On the circular arc from (a, 0, 0) to (0, a, 0), \rho= a, \theta goes from 0 to \pi/2a, and \phi= \pi. On the straight line from (0, a, 0), \rho goes from a to 0, \theta= \pi/2, and \phi= \pi. That should be all you need.

As far as the surface integral is concerned, the path, and so the surface bounded by it, is entirely in the xy-plane so \phi= 0 for every point. You can write the "position vector" in terms of \rho and \theta as \vec r= \rho \vec\rho+ \theta\vec\theta+ \pi\vec\phi.