# Thread: Calculate div F and curl F (polar coordinates)

1. ## Calculate div F and curl F (polar coordinates)

F(r,θ) = ri + sinθj, where (r,θ) are polar coordinates in the plane.

In my book are given both formulas to calculate div F and curl F - but only for cartesian coordinates. What counts for polar coordinates in the plane? It is like I am to differentiate ri with respect to x and sin
θj with respect to y: Am I wrong to think so?

Best regards

2. ## Re: Calculate div F and curl F (polar coordinates)

Your $\mathbf{F}$ is already in terms of $\mathbf{i}$ and $\mathbf{j}$, so you just need to convert $r$ and $\sin{\theta}$ to Cartesian coordinates. So $\mathbf{F}(x,y)=\sqrt{x^2+y^2}\mathbf{i}+\frac{y}{ \sqrt{x^2+y^2}}\mathbf{j}$.

- Hollywood

3. ## Re: Calculate div F and curl F (polar coordinates)

div F equals cos θ (1 + (1/r) cos θ)
If I take the derivative of i with respect to x and plus it with the derivative j with respect to y I don't get this answer above (taken from the book).

Best regards Kaemper

4. ## Re: Calculate div F and curl F (polar coordinates)

Originally Posted by kaemper
div F equals cos θ (1 + (1/r) cos θ)
If I take the derivative of i with respect to x and plus it with the derivative j with respect to y I don't get this answer above (taken from the book).

Best regards Kaemper
I get div F is $\frac{x \left(x^2+x+y^2\right)}{\left(x^2+y^2\right)^{3/2}}$

and your expression above converted to cartesian coordinates is

$\frac{x \left(x^2+x+y^2\right)}{\left(x^2+y^2\right)^{3/2}}$

and you can see the two are identical

5. ## Re: Calculate div F and curl F (polar coordinates)

I got it, thank you.