Finding the curl and divergence

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    Curl Divergence
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SUMMARY

The curl and divergence of the vector field \(\vec{F}(x,y,z) = x^2y\vec{i} + y^2z^3\vec{j} + xyz\vec{k}\) have been calculated correctly. The curl is \((xz - 3y^2z^2)\vec{i} + (-yz)\vec{j} + (-x^2)\vec{k}\) and the divergence is \(2xy + 2yz^3 + xy\), which can also be expressed as \(3xy + 2yz^3\). The calculations were confirmed by participants in the discussion.

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Homework Statement



\vec{F}(x,y,z) = x^2y\vec{i} + y^2z^3\vec{j} + xyz\vec{k}

Homework Equations



The Attempt at a Solution



I got:

Curl: (xz - 3y^2z^2)\vec{i} + (-yz)\vec{j} + (-x^2)\vec{k}
Div: 2xy + 2yz^3 + xy

Are these right?
 
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duki said:
[
Curl: (xz - 3y^2z^2)\vec{i} + (-yz)\vec{j} + (-x^2)\vec{k}
Div: 2xy + 2yz^3 + xy

Are these right?

Looks good to me (although I would write the divergence as 3xy + 2yz^3:wink:)
 
Awesomes. Thanks
 

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