line integral of $\frac{x dx + ydy }{\sqrt{x^2 + y^2} } $ along any path inside the annulus joining two concentric circles (viewed as shells)

Question

Find $\int \frac{x dx + ydy }{\sqrt{x^2 + y^2} } $ along any path inside the shell joining the concentric shells $x^2 + y^2 = R_1^2 $ and $x^2 + y^2 = R_2^2$, where $R_1 < R_2$

To find the line integral, I choose my path as $$\alpha(r,\theta) = (r, \theta_0),$$ for some $\theta_0\in [0, 2\pi]$.

so for $$f(r,\theta) = (cos\theta, sin\theta)$$

we have

$$ d\alpha(r,\theta) = (dr, 0) \\ \int f d\alpha = \int_{R_1}^{R_2} cos\theta_0 dr = cos(\theta_0) (R_2 - R_1),$$ which means the line integral does depend on the angle that we choose, but the answer provided to us claims that the answer is $$R_2 - R_1$$, so my question is that is there anything wrong in my calculations ?

Answer 1

So you chose the path to be with a constant angle, meaning $$\theta(t)=\theta_0 , r(t)=R_1+(R_2-R_1)t$$ $$\vec{r}(t) = r(t)\cdot (\cos \theta_0 , \sin \theta_0) $$ $$ \vec{dr} = \frac{dr}{dt} dt (\cos \theta_0 , \sin \theta_0) = (R_2-R_1)(\cos \theta_0 , \sin \theta_0)$$ and so, $$\vec{f}\cdot \vec{dr} = (R_2-R_1)dt $$ And the integral will be $$\int_0^1 (R_2-R_1) dt = R_2-R_1$$ The answer doesn't depend on $\theta_0$!

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Moreover, you can calculate the potential $\phi$ creating $\vec{f}$ and by the gradient theorem, $$\phi(b)-\phi(a) = \int \nabla \phi \cdot \vec{dr} = \int \vec{f}\cdot \vec{dr}$$

In polar coordinates, $$\vec{f} = \hat{r}$$ and so $$\phi(r,\theta)=r$$ meaning each circle is an equai-potential - again confirming the answer doesn't depend on the angle.

Answer 2

Since $xdx+ydy=d(r^2)/2=rdr$, your integral is $\int_{R_1}^{R_2}dr$, which is the desired answer.