I am supposed to compute the line integral along the path described in the picture using spherical coordinates. When computing the last path$(0,1,2) \to (0,0,0)$, I believed that there were infinitesimal changes $dr$ and $d\theta$(because the path goes from a point in yz plane back on the xy plane, where $\theta = \frac{\pi}{2}$), but in the solution, there was only a $dr$ involved.
The spherical coordinate I used here is $(r, \theta, \phi)$, where $\theta \in [0, \pi], \phi \in [0, 2\pi]$.
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Is this Div grad curl and all that? – tryst with freedom May 20 '22 at 14:44
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I mean just plug in the formula and check. Is there any reason why you're trying to check intuitioN? – tryst with freedom May 20 '22 at 14:44
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I want to know if I could tell from the given graph that $\theta$ changes – rorochichichi May 20 '22 at 14:47
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Check how your book defines $\theta$ and $\ohi$. – tryst with freedom May 20 '22 at 14:49
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$\theta$ is the polar angle down from the z axis while $\phi$ is the azimuth angle around the z axis that starts from x axis – rorochichichi May 20 '22 at 14:51
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Yeah you're sliding down the radius along a line of constant $\theta$ aren't you? Try compare this pic with the standard spherical geometry pic – tryst with freedom May 20 '22 at 14:53
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Thank you for this confirmation! That makes me more confident about my answer, because mine disagreed with the given solution. – rorochichichi May 20 '22 at 14:56
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In spherical coordinates the origin can be defined just by $r=0$. $\theta$ and $\phi$ have no meaningful values at this point, since the position vector of origin is null vector and null vector can have any direction or say no direction.
Now, coming to the last integral from $(0,1,2) \rightarrow (0,0,0)$ , notice that the $\theta$ remains constant along the whole path, except at the origin where it has no particular (or well defined) value, and $\phi=\frac{\pi}{2}$ is also constant. So the integral is calculated by considering $dr$ only.
