# Line integral of a spherical vector field over cartesian path

## Homework Statement

Compute the line integral of

$\vec{v} = (rcos^{2}\theta)\widehat{r} - (rcos\theta sin\theta)\widehat{\theta} + 3r\widehat{\phi}$

over the line from (0,1,0) to (0,1,2) (in Cartesian coordinates)

## The Attempt at a Solution

Well, I expressed the path as a parametrized vector

$\vec{r}(t) = \frac{1}{sint} \widehat{r} + t\widehat{\theta} + \frac{\pi}{2}$, t:(arctan(1/2), pi/2)

the derivative of which is

$\vec{r}'(t) = -\frac{cost}{sin^{2}t} \widehat{r} + \widehat{\theta}$

I'm looking for the integral to be equal to 2, but whenever I work it out I get a mess of logarithms and square roots. Have I parameterized this the wrong way?

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tiny-tim
Homework Helper
Hi PeteyCoco!
… over the line from (0,1,0) to (0,1,2) (in Cartesian coordinates)
Wouldn't it be massively easier to use the parameter z ?

(and convert v to Cartesian)

Hi PeteyCoco!

Wouldn't it be massively easier to use the parameter z ?

(and convert v to Cartesian)
Yeah, it was much simpler. Should I be able to get exactly the same answer if I work it out in terms of spherical coordinates though? When I did the arithmetic on the answer from that process I got something around 1.975 instead of 2. Is that due to poor math on my part or is the result off because of the unnatural fit of a line in spherical coordinates?

tiny-tim
Homework Helper
Hi PeteyCoco!

(just got up :zzz:)
Should I be able to get exactly the same answer if I work it out in terms of spherical coordinates though?
Yes, but you have to be very careful about the line element in spherical coordinates (I expect that's where you went wrong) …

in Cartesian corordinates, it's just dz, with no factors !!

If you want us to find your mistake (the answer should have been exactly the same), you'll have to type it out for us.

vanhees71
The $z$-axis is a coordinate singularity in standard spherical coordinates!