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Math Help - another vector calc question

  1. #1
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    another vector calc question

    curl(ФA) at the point (1,1,1)
    where A is a vector=(2xz^2)i-(yz)j+(3xz^3)k
    and Ф=(x^2)yz

    i first multiplied each component of A by (x^2)yz
    i then did del cross the vector i got after multiplying A by the scalar to figure out the curl
    i came up with (3x^3*z^4+2x^2*y^2*z)i+(9x^2*y*z^4-3x^3*y*z^2)j+(-2y^2*z^2*x-x^3*z^3)k
    i then plugged in the point (1,1,1) to the equation for the curl vector and i got 5i+6j-3k but the answer in the book was 5i-3j-4k
    am i on the right track here?
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  2. #2
    MHF Contributor
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    When I did the curl, I obtained (3{x}^{3}{z}^{4}+2{x}^{2}yz)i-(9{x}^{2}y{z}^{4}-2x{z}^{3})j+(-2x{y}^{2}z-2{x}^{3}{z}^{3})k which when (1,1,1) is plugged in arrives at the correct answer.
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