I would like to compute the mean value of a second order tensor $\mathbf{T}$ expressed in planar cylindrical coordinates.
The mean value for any second order tensor is (reference [1] page 101) $$\langle\mathbf{T}\rangle=\frac{1}{\text{vol}\left[B\right]}\int_{B}\mathbf{T}\mathrm{d}V$$
where $B$ is a region in Euclidean space and $\mathrm{d}V$ a volume element in the appropriate coordinate system. I would like to evaluate this for a tensor $$\mathbf{T}=T_{rr}\left(r\right)\mathbf{e}_{r}\left(\varphi\right)\otimes\mathbf{e}_{r}\left(\varphi\right)+T_{\varphi\varphi}\left(r\right)\mathbf{e}_{\varphi}\left(\varphi\right)\otimes\mathbf{e}_{\varphi}\left(\varphi\right)$$
which takes diagonal form in the planar cylindrical coordinate system. The radial basis vector $\mathbf{e}_{r}$ and the circumferential (tangential) basis vector $\mathbf{e}_{\varphi}$ are
$$\mathbf{e}_{r}\left(\varphi\right)=\left(\cos\varphi\right)\mathbf{e}_{x}+\left(\sin\varphi\right)\mathbf{e}_{y}\qquad\mathbf{e}_{\varphi}\left(\varphi\right)=\left(-\sin\varphi\right)\mathbf{e}_{x}+\left(\cos\varphi\right)\mathbf{e}_{y}$$
where $\left\{ \mathbf{e}_{x},\mathbf{e}_{y}\right\}$ is the cartesian basis. The inverse transformation is
$$\mathbf{e}_{x}=\left(\cos\varphi\right)\mathbf{e}_{r}-\left(\sin\varphi\right)\mathbf{e}_{\varphi}\qquad\mathbf{e}_{y}=\left(\sin\varphi\right)\mathbf{e}_{r}+\left(\cos\varphi\right)\mathbf{e}_{\varphi}$$
Since I am integrating over an area (planar cylindrical coordinates), the volume element is $r\mathrm{d}r\mathrm{d}\varphi$. I integrate over a disk from $a$ to $b$, evaluating $\langle\mathbf{T}\rangle$ as $$\langle\mathbf{T}\rangle=\frac{1}{\pi\left(b^{2}-a^{2}\right)}\int_{r=a}^{b}\int_{\varphi=0}^{2\pi}T_{rr}\left(r\right)\mathbf{e}_{r}\left(\varphi\right)\otimes\mathbf{e}_{r}\left(\varphi\right)+T_{\varphi\varphi}\left(r\right)\mathbf{e}_{\varphi}\left(\varphi\right)\otimes\mathbf{e}_{\varphi}\left(\varphi\right)r\mathrm{d}r\mathrm{d}\varphi$$
The $r$ and $\varphi$ integration are decoupled, $$\begin{aligned}\pi\left(b^{2}-a^{2}\right)\langle\mathbf{T}\rangle= & \left(\int_{r=a}^{b}T_{rr}\left(r\right)r\mathrm{d}r\right)\underbrace{\left(\int_{\varphi=0}^{2\pi}\mathbf{e}_{r}\left(\varphi\right)\otimes\mathbf{e}_{r}\left(\varphi\right)\mathrm{d}\varphi\right)}_{I_{r}}\\ & +\left(\int_{r=a}^{b}T_{\varphi\varphi}\left(r\right)r\mathrm{d}r\right)\underbrace{\left(\int_{\varphi=0}^{2\pi}\mathbf{e}_{\varphi}\left(\varphi\right)\otimes\mathbf{e}_{\varphi}\left(\varphi\right)\mathrm{d}\varphi\right)}_{I_{\varphi}} \end{aligned}$$
I would now like to focus on the integral $I_{r}$ I rewrite $\mathbf{e}_{r}\left(\varphi\right)\otimes\mathbf{e}_{r}\left(\varphi\right)$ in the cartesian basis, $$\begin{aligned}I_{r}= & \int_{\varphi=0}^{2\pi}\left(\cos\varphi\right)^{2}\mathbf{e}_{x}\otimes\mathbf{e}_{x}+\left(\sin\varphi\right)^{2}\mathbf{e}_{y}\otimes\mathbf{e}_{y}\\ & +\left(\cos\varphi\right)\left(\sin\varphi\right)\mathbf{e}_{x}\otimes\mathbf{e}_{y}+\left(\cos\varphi\right)\left(\sin\varphi\right)\mathbf{e}_{y}\otimes\mathbf{e}_{x}\mathrm{d}\varphi \end{aligned} $$
After evaluating this integral, I obtain $$I_{r}=\pi\left(\mathbf{e}_{x}\otimes\mathbf{e}_{x}+\mathbf{e}_{y}\otimes\mathbf{e}_{y}\right)$$
which after applying the inverse transformation back to cylindrical coordinates becomes $$I_{r}=\pi\left(\mathbf{e}_{r}\left(\varphi\right)\otimes\mathbf{e}_{r}\left(\varphi\right)+\mathbf{e}_{\varphi}\left(\varphi\right)\otimes\mathbf{e}_{\varphi}\left(\varphi\right)\right)$$
There is something that feels deeply wrong about this, as I integrated out the $\varphi$ -dependence in cartesian coordinates but re-introduced it through the inverse transformation. Also, from a continuum mechanics point of view, the result seems wrong. Coud you please point out to me where I go wrong and how I can evaluate this integral correctly?
[1] Gonzalez, Oscar, and Andrew M. Stuart. A first course in continuum mechanics. Cambridge University Press, 2008.