Problems with conversions to spherical coordinates involving a line integral

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SUMMARY

The discussion focuses on calculating the line integral of the vector field A = 4r + 3θ - 2φ around a closed path in spherical coordinates. The path consists of a straight line along the x-axis, a circular arc, and another straight line returning to the origin. The user seeks clarification on converting Cartesian coordinates to spherical coordinates for the line integral and determining the differential element ds. Additionally, the surface integral of the curl of A over the enclosed area is to be compared with the line integral results.

PREREQUISITES
  • Understanding of spherical coordinates and their conversion from Cartesian coordinates.
  • Familiarity with vector calculus, specifically line and surface integrals.
  • Knowledge of the curl operation and its application in vector fields.
  • Proficiency in using integral notation and differential elements in multiple dimensions.
NEXT STEPS
  • Study the conversion formulas between Cartesian and spherical coordinates.
  • Learn about the differential line element ds in spherical coordinates.
  • Explore the properties of curl and its implications in vector fields.
  • Practice solving line and surface integrals in various coordinate systems.
USEFUL FOR

Students and professionals in physics and engineering, particularly those working with vector calculus and integrals in spherical coordinates.

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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?
 
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HallsofIvy said:
On the straight line from (0,0,0) to (a, 0, 0), [itex]\rho[/itex] goes from 0 to a, [itex]\theta= 0[/itex] and [itex]\phi= \pi[/itex]. On the circular arc from (a, 0, 0) to (0, a, 0), [itex]\rho= a[/itex], [itex]\theta[/itex] goes from 0 to [itex]\pi/2a[/itex], and [itex]\phi= \pi[/itex]. On the straight line from (0, a, 0), [itex]\rho[/itex] goes from a to 0, [itex]\theta= \pi/2[/itex], and [itex]\phi= \pi[/itex]. 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 [itex]\phi= 0[/itex] for every point. You can write the "position vector" in terms of [itex]\rho[/itex] and [itex]\theta[/itex] as [itex]\vec r= \rho \vec\rho+ \theta\vec\theta+ \pi\vec\phi[/itex].
 

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