Curl of Vector Field u = yi+(x+z)j+xy^(2)k: Step-by-Step Calculation Method

andrey21
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Find the curl of the following vector field

u = yi+(x+z)j+xy^(2)k

Now using the method I've bin taught similar to finding determinant of 3x3 matrix here is my answer
i(2yx-1) -j(y^2) +k(0)Just looking for confirmation if this is correct or any basic errors I have made thank you.
 
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Confirmed that it is correct, then.
 
Thank you Char.limit just a follow up question I am asked:

Find (curl u).v

Where
v = xi+(y^(2) - 1)j+(1-x^(2))k

Is that just the dot product of the vector v and the curl established for u. Thank you
 
Yes it is. And you can just ignore the k part completely.
 
So would that give me:

(2x^(2) y -x) +(y^(4) - y^(2))

Also do I still need the i j k notations??
 
YOu have obtained a scalar. There's no more unit vector involved.
 
Oh ok so my final answer would just be 2x^(2) - x +y^(4) - y^(2)

correct?
 
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