# Proving vector calculus relationships

• February 7th 2013, 10:28 PM
Trianagt
Proving vector calculus relationships
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
• February 7th 2013, 10:46 PM
jakncoke
Re: Proving vector calculus relationships
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 $\begin{bmatrix} 0\\0\\0 \end{bmatrix}$
• February 8th 2013, 05:33 AM
HallsofIvy
Re: Proving vector calculus relationships
Trianagt, to use jakncokes' answer you will also need to know that $\nabla\times (f\vec{v})= \nabla f\times v+ f\nabla\times v$