Proving vector calculus relationships

**Trianagt** · February 7th 2013, 10:28 PM

How would you go about proving:

Curl(f(r)R) where R is the position vector x^(2)i +y^(2)j +z^(2)k

I know it should equal zero but once Ihave it in its components I am struggling to make them equal zero.

Thank you

**jakncoke** · February 7th 2013, 10:46 PM

so basically the curl is evaluates to $\begin{bmatrix}\frac{\partial}{\partial{y}}z^2 - \frac{\partial}{\partial{z}}y^2 \\ \frac{\partial}{\partial{x}}z^2 - \frac{\partial}{\partial{z}}x^2 \\ \frac{\partial}{\partial{x}}y^2 - \frac{\partial}{\partial{y}}x^2 \end{bmatrix}$ which is clearly $\begin{bmatrix} 0\\0\\0 \end{bmatrix}$

**HallsofIvy** · February 8th 2013, 05:33 AM

Trianagt, to use jakncokes' answer you will also need to know that $\nabla\times (f\vec{v})= \nabla f\times v+ f\nabla\times v$

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