check for incompressibility in sin and cos

Question

Given $\mathbf{u}=\sin{x}\sin{y}\mathbf{i}+\cos{x}\cos{y}\mathbf{j}$ check that it satisfies the condition:

$\nabla\cdot\mathbf{u}=0$

I tried approaching it likeso:

$\frac{dx}{dy}=\sin{x}\sin{y}\mathbf{i}+\cos{x}\cos{y}\mathbf{j}=\frac{dx}{\sin{x}\cos{x}}=\frac{dy}{\cos{x}+\sin{y}}$

Am I calculating this correctly? I'm trying to check whether $\mathbf{u}$ is incompressible?

Answer 1

Given $\mathbf{u}=\sin{x}\sin{y}\mathbf{i}+\cos{x}\cos{y}\mathbf{j}$ check that it satisfies the condition: $\nabla\cdot\mathbf{u}=0$.

Note that $$\color{blue}{\boxed{\nabla:=\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right)}}$$ and $$\mathbf{u}=(\sin(x)\sin(y),\cos(x)\cos(y))$$ Then, \begin{eqnarray*} \nabla\cdot \mathbf{u}&=&\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right)\cdot (\sin(x)\sin(y),\cos(x)\cos(y))\\ &=&\frac{\partial}{\partial \color{red}{x}}\left(\color{red}{\sin(x)}\sin(y) \right)+\frac{\partial}{\partial \color{red}{y}}\left(\cos(x)\color{red}{\cos(y)} \right)\\ &=&\cos(x)\sin(y)\color{red}{-}\sin(y)\cos(x)\\ &=&0 \end{eqnarray*}

Therefore, $\nabla \cdot \mathbf{u}=0$.