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Thread: Find Curl of Vector Field

  1. #1
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    Find Curl of Vector Field

    Find the curl of vector field.

    F(x, y, z) = xy i + yz j + xy k

    ∇xF = x^2 z i + y^2x j + z^2 k

    The book's answer is

    ∇xF = -y i - z j - x k

    Who is right and why?
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    Re: Find Curl of Vector Field

    Quote Originally Posted by USNAVY View Post
    Find the curl of vector field.

    F(x, y, z) = xy i + yz j + xy k

    ∇xF = x^2 z i + y^2x j + z^2 k

    The book's answer is

    ∇xF = -y i - z j - x k

    Who is right and why?

    Show some effort on your part.
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  3. #3
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    Re: Find Curl of Vector Field

    I actually have a suspicion that there is a typo in the question. I think the last term of the vector field should be xz k instead of xy k. When it is xz k, then my answer agrees with the textbook.
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    Re: Find Curl of Vector Field

    Quote Originally Posted by USNAVY View Post
    I actually have a suspicion that there is a typo in the question. I think the last term of the vector field should be xz k instead of xy k. When it is xz k, then my answer agrees with the textbook.
    Well find $\nabla F$ and show it. Then explain how it differs from the text.
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  5. #5
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    Re: Find Curl of Vector Field

    The book's answer is correct. I can't imagine how you got your answer! Your answer has polynomials of higher power than the original vector while differentiating always reduces the power of a polynomial.

    \nabla\times F= \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} & \frac{\partial F}{\partial z} \\ xy & yz & xz \end{array}\right|= (0- x)\vec{i}-(z- 0)\vec{j}+ (0- x)\vec{k}= -x\vec{i}- \vec{j}- \vec{k}.
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