Undergrad Integration over a part of a spherical shell in Cartesian coordinates

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The discussion centers on integrating the vector function f = -xj + yi over a portion of a spherical shell with radius R, specifically the projection of a circle of radius b in the x-y plane. The poster is exploring the feasibility of performing this integration in Cartesian coordinates, as they are familiar with the integration intervals for x and y. A suggestion is made to use the surface integral approach with the area element dA = sec(γ) dx dy, where γ is the angle between the surface normal and the z-axis. Clarification is sought on whether the integration is for a surface or volume, with a note that if it involves a thickness Δr, the surface integral result could be multiplied by Δr. The discussion highlights the challenges of setting integration limits in spherical coordinates for this specific problem.
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I am modeling some dynamical system and I came across integral that I don't know how to solve. I need to integrate vector function f=-xj+yi (i and j are unit vectors of Cartesian coordinate system).
I need to integrate this function over a part of spherical shell of radius R. This part is actually projection of circle (that lies in x-y plane) of radius 'b' on this sphere. I know that it's usually easier to integrate spherical shell in spherical coordinates but it's really tricky to set the intervals of integration in this problem in spherical coordinates.
-So my question is, is it possible to integrate this in Cartesian coordinates. Because i know integration intervals: I just need to set the interval for x and y as circle in x-y plane (z interval would be little tricky)

I also posted image so you can better see what the problem is...
integration.jpg
 

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It can be integrated over ## dx \, dy ## with ## dA=\sec{\gamma} \, dx \, dy ##, where ## \gamma ## is the angle between the normal to the surface at any point and the z-axis. ## \\ ## It has been quite a number of years since I did such an integral in an Advanced Calculus course, but I still remember doing it this way. ## \\ ## Edit: I may have misread this: Are you doing a surface integral, or an integral over some volume? What I did is for a surface integral. If it is over a thickness of ## \Delta r ## , you might be able to multiply the surface integral result by ## \Delta r ## . ## \\ ## Additional edit: I believe in this case ## \cos{\gamma}=\frac{z}{R} ##.
 
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