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Math Help - Calculate div F and curl F (polar coordinates)

  1. #1
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    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
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  2. #2
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    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
    Thanks from kaemper
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  3. #3
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    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
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  4. #4
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    Re: Calculate div F and curl F (polar coordinates)

    Quote Originally Posted by kaemper View Post
    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
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  5. #5
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    Re: Calculate div F and curl F (polar coordinates)

    I got it, thank you.
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