Applying gradient in spherical coordinates to vector in cartesian coordinates [closed]

I am trying to calculate the gradient of a vector field $$\boldsymbol{u}$$.

In cartesian coordinates, I would normally do

$$\left(\nabla\boldsymbol{u}\right)_{ij}=\partial_{i}u_{j}=\left(\begin{array}{ccc} \partial_{x}u_{x} & \partial_{x}u_{y} & \partial_{x}u_{z}\\ \partial_{y}u_{x} & \partial_{y}u_{y} & \partial_{y}u_{z}\\ \partial_{z}u_{x} & \partial_{z}u_{y} & \partial_{z}u_{z} \end{array}\right)$$

but I want to express the gradient in spherical coordinates, i.e.

$$\nabla\boldsymbol{u}=\left(\hat{\boldsymbol{r}}\partial_{r}+\frac{\hat{\theta}}{r}\partial_{\theta}\frac{\hat{\phi}}{r\sin\theta}\partial_{\phi}\right)\boldsymbol{u}$$

However, $$\boldsymbol{u}$$ is defined as $$u_i = M_{ij} f_j$$, in cartesian components. $$M=M(x,y,z)$$ is a complicated expression, whereas $$f$$ is constant. So I should convert $$u$$ to spherical coordinates by transforming both $$M$$ and $$f$$ first, then apply the gradient $$\nabla$$ in spherical coordinates. This is quite laborious and in the end not crucial to my end goal, which is to integrate $$\nabla \boldsymbol{u}$$ over a spherical surface.

So my question is, could I leave $$\boldsymbol{u}$$ in cartesian coordinates and apply the gradient in spherical coordinates to it? This would look something like

$$\left(\nabla\boldsymbol{u}\right)_{11}=\partial_{r}u_{x}$$ $$\left(\nabla\boldsymbol{u}\right)_{12}=\partial_{r}u_{y}$$ $$\left(\nabla\boldsymbol{u}\right)_{22}=r^{-1}\partial_{\theta}u_{x}$$ $$\left(\nabla\boldsymbol{u}\right)_{22}=r^{-1}\partial_{\theta}u_{y}$$ $$\text{etc.}...$$

Would this present problems when integrating over a sphere, e.g.

$$\iint_{S}\nabla\boldsymbol{u}\cdot\boldsymbol{dS}\ \ ?$$

How should $$\boldsymbol{dS}$$ be rewritten in order for this hybrid formulation to make sense?

For clarification, write $$\vec{u} = u_x \hat{x} + u_y \hat{y} + u_z\hat{z}$$ in components, and $$d\vec{S} = \hat{r} R^2 \sin\theta d\theta d\phi$$ $$\vec{I} \equiv \oint \oint \vec{\nabla} \vec{u} \cdot d\vec{S}$$ \begin{align} I_\alpha = & \oint \oint \vec{\nabla} u_\alpha(\vec{r})\cdot d\vec{S}\\ =& R^2 \int_0^\pi \sin\theta d\theta \int_0^{2\pi} d\phi \frac{\partial u_\alpha(\vec{r})}{\partial r}\\ =& R^2 \int_0^\pi \sin\theta d\theta \int_0^{2\pi} d\phi \left\{\frac{\partial x}{\partial r} \frac{\partial u_\alpha(\vec{r})}{\partial x} +\frac{\partial y}{\partial r} \frac{\partial u_\alpha(\vec{r})}{\partial y} + \frac{\partial z}{\partial r} \frac{\partial u_\alpha(\vec{r})}{\partial z} \right\}\\ =& R^2 \int_0^\pi \sin\theta d\theta \int_0^{2\pi} d\phi \left\{\sin\theta\cos\phi \frac{\partial u_\alpha(\vec{r})}{\partial x} +\sin\theta\sin\phi \frac{\partial u_\alpha(\vec{r})}{\partial y} + \cos\theta \frac{\partial u_\alpha(\vec{r})}{\partial z} \right\} \end{align}

Where we use \begin{align} x = & r \sin\theta\cos\phi;\\ y = & r \sin\theta\sin\phi;\\ z = & r \cos\theta. \end{align}

• I see.. However, I can't see how this formalism would apply to, say, the integral of $\left(\nabla\boldsymbol{u}\right)^{T}$, or $\mathbb{I}$ (identity matrix) Apr 7, 2021 at 17:59
• Also, why is there a sum over $\alpha$ in your integrals? Did you mean $I_\alpha = ...$ instead? Apr 7, 2021 at 18:02
• "why is there a sum over α "? I worte $\vec{u} = \sum_a \hat{a} u_a$. The "$\cdot$" is operate between $\vec{\nabla}$ and $d\vec{S}$.
– ytlu
Apr 7, 2021 at 18:16
• If you cannot tell how to decompose, just write all in indices form.
– ytlu
Apr 7, 2021 at 18:17
• Let me edit to a form your feel confortable.
– ytlu
Apr 7, 2021 at 18:23