how to calculate curl((r_)/r^3)?

the r_ inside the braket means vector r,I dont know how to write it proper.

Some onw please help!

4 Answers

  • kb
    Lv 7
    10 years ago
    Favorite Answer

    r = <x, y, z>, ||r|| = (x^2 + y^2 + z^2)^(-1/2)

    So, we need to compute

    curl <x(x^2 + y^2 + z^2)^(-1/2), y(x^2 + y^2 + z^2)^(-1/2), z(x^2 + y^2 + z^2)^(-1/2)>

    = < -yz(x^2 + y^2 + z^2)^(-3/2) + -yz(x^2 + y^2 + z^2)^(-3/2),

    -zx(x^2 + y^2 + z^2)^(-3/2) + zx(x^2 + y^2 + z^2)^(-3/2),

    -xy(x^2 + y^2 + z^2)^(-3/2) + xy(x^2 + y^2 + z^2)^(-3/2)>

    = <0, 0, 0>.

    I hope this helps!

  • 5 years ago

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  • Anonymous
    10 years ago

    I got the same except for the II r II is II r II^3 so wouldnt it be (x^2 + y^2 + z^2)^-3/2 to start with and then (x^2 + y^2 + z^2)^-5/2 when differentiated?? The answer comes out to the same <0,0,0> but working is slightly different? Is this right? :)

    Source(s): my brain
  • 10 years ago


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