Vector Calculus: Computing (V•∇)U and (U•∇)V with Given Functions

[CalcYouLater]

Homework Statement

Hello. I want to see if I am interpreting the following correctly, I certainly don't expect anyone to work the problem out as it is (at least with my approach) fairly tedious.

Compute the following:

(\vec{V}\cdot\nabla)\vec{U}
(\vec{U}\cdot\nabla)\vec{V}

Given:

\vec{r} = x\hat{x}+y\hat{y}+z\hat{z}
\vec{V}=yz\hat{x}+xz\hat{y}+xy\hat{z}

\vec{U}=\frac{x^2+y^2+z^2}{(x^2+y^2)^{\frac{3}{2}}}\hat{z}-\frac{z}{(x^2+y^2)^{\frac{3}{2}}}\vec{r}

The Attempt at a Solution

Here is my approach

(\vec{U}\cdot\nabla)\vec{V}= U_{x}\frac{d}{dx}(yz\hat{x}+xz\hat{y}+xy\hat{z})+U_{y}\frac{d}{dy}(yz\hat{x}+xz\hat{y}+xy\hat{z})+U_{z}\frac{d}{dz}(yz\hat{x}+xz\hat{y}+xy\hat{z})

I evaluate the first part of the expression above as:

U_{x}\frac{d}{dx}(yz\hat{x}+xz\hat{y}+xy\hat{z})=\frac{-z^{2}x}{(x^2+y^2)^{\frac{3}{2}}}\hat{y}+\frac{-zxy}{(x^2+y^2)^{\frac{3}{2}}}\hat{z}

After doing the same for the rest of the equation, and adding the components, my result is a vector.

I use the same approach for:

(\vec{V}\cdot\nabla)\vec{U}

Does anyone disagree? If not, does anyone see a more efficient way of doing this?

 

[Dick]

Without checking everything, it looks like you are on the right track. And I don't see any easier way to do it. It looks like it was set up as a deliberately tedious exercise. I hate those.