Maxwell stress tensor missing angular factor

[Another]

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my think

if $\hat{r} = \sin(θ) \cos( φ) \hat{x} +\sin(θ) \sin( φ) \hat{y} +\cos(θ) \hat{z}$
$da = R^2 \sin(θ) dθdφ \hat{r} = da_{x} \hat{x} + da_{x} \hat{y} + da_{z} \hat{z}$
So
$da_{x} = R^2 \sin^2(θ) \cos(φ) dθdφ$
$da_{y} = R^2 \sin^2(θ) \sin(φ) dθdφ$
$da_{z} = R^2 \sin(θ) \cos(θ) dθdφ$
where
$\int_{0}^{2π} \cos(φ) \,dφ =\int_{0}^{2π} \sin(φ) \,dφ = 0$

$\overleftrightarrow{T} ⋅ da = 0 + 0 + T_{zz} ⋅ da_{z}$

$T_{zz} ⋅ da_{z} = \frac{ε_{0}}{2} \left(\frac{Q}{4πε_{o}R^2}\right)^2(\cos^2(θ) - \sin^2(θ)) ⋅ R^2\sin(θ) \cos(θ) dθdφ$

But why in textbook give by
$\frac{ε_{0}}{2} \left(\frac{Q}{4πε_{o}R^2}\right)^2 ⋅ R^2\sin(θ) \cos(θ) dθdφ = \frac{ε_{0}}{2} \left(\frac{Q}{4πε_{o}R}\right)^2\sin(θ) \cos(θ) dθdφ$

where are $(\cos^2(θ) - \sin^2(θ))$ ?. $(\cos^2(θ) - \sin^2(θ))$ is missing .
I don't understand

 

[Orodruin]

The x and y components do not vanish. There is a $\varphi$ dependence also in $T_{zx}$ and $T_{zy}$ that you need to take into account. You cannot just integrate the area element assuming that they are constant.