- #1
flyingpig
- 2,579
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Homework Statement
Find the curl of the vector field
[tex]\mathbf{F} = <xyz,0,-x^2 y>[/tex]
The Attempt at a Solution
I am mostly just having problems with computing the determinant. I could just go with crossing the first row and first column. But i noticed that the intermediate step
[tex]\begin{vmatrix}
\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\
0& -x^2 y
\end{vmatrix}\hat{i}[/tex]
Like when we compute the 2 x 2 determinant, we multiply the elements together. But my question is that why we do take granted that multiplying the partial with respect to y with -x2y is equal to taking the partial of -x2y with respect to y?
Also, I tried crossing out from the bottom, where the 0 is to make the determinant easier and I got a different answer. I also tried using the Rule of Sarrus for this 3 x 3 matrix and I also got a different answer.
The one with the Rule of Sarrus is a bit long to type out, so i will only show the one where I crossed with the 0 (I think it's called the co-factor expansion or something)
So crossing the bottom row and first column I get
[tex]\begin{vmatrix}
\hat{i} & \hat{j} &\hat{k} \\
\frac{\partial }{\partial x} &\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\
xyz & 0 & -x^2 y
\end{vmatrix} = xyz\begin{vmatrix} \hat{j} & \hat{k}\\\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\end{vmatrix} - x^2 y \begin{vmatrix} \hat{i}&\hat{j}\\\frac{\partial }{\partial x}&\frac{\partial }{\partial y}\end{vmatrix} = xyz\left(\frac{\partial\hat{j} }{\partial z}- \frac{\partial \hat{k}}{\partial y}\right) -x^2 y \left ( \frac{\partial \hat{i}}{\partial y}- \frac{\partial \hat{j}}{\partial x} \right )[/tex]
The Solution
[PLAIN]http://img641.imageshack.us/img641/1914/unledzw.jpg
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