# Evaluating the components of Maxwell's stress tensor

I was going through the Maxwell's stress tensor section of Introduction to Electrodynamics by Griffiths. In the example 8.2(screenshot below), I fail to understand how the equation 8.23 (in the image) gives, $$\frac{\epsilon_{0}}{2}\big(\frac{Q}{4\pi{\epsilon_0} r}\big)^2 \sin\theta \cos\theta \,d{\theta}\,d{\phi}.$$ If we use the equation $$\hat{r}=\sin{\theta} \cos{\phi} \hat{x}+ \sin\theta \sin\phi \hat{y} + \cos\theta \hat{z}$$ in $$da=R^2 \sin\theta \,d\theta \,d\phi \hat{r},$$ and substitute it in $\big(T\cdot da \big)_z$ then it gives a different answer. Can somebody elaborate on the steps linking $\big(T\cdot da \big)_z$ and $$\frac{\epsilon_{0}}{2}\big(\frac{Q}{4\pi{\epsilon_0} r}\big)^2 \sin\theta \cos\theta \,d{\theta}\,d{\phi}~?$$

PS: I really apologize if the question seems too silly but I have spent many hours wondering about this. I also do not have a great background in Tensors.

• If you want your work to be checked for a clumsy mistake, write it out; it's impossible to tell where you might have went wrong. – knzhou Mar 13 '16 at 23:43

$$(\mathbf T \cdot \mathbf{da})_z = \epsilon_0 \left( \frac{Q}{4\pi \epsilon_0 R^2} \right)^2 R^2 \sin \theta \,d \theta d \phi \left[ \sin^2 \theta \cos^2 \phi \cos \theta + \sin^2 \theta \sin^2 \phi \cos \theta + \frac{\cos \theta}{2} (\cos^2 \theta - \sin^2 \theta) \right] = \epsilon_0 \left( \frac{Q}{4\pi \epsilon_0 R^2} \right)^2 R^2 \sin \theta \,d \theta d \phi \left[ \sin^2 \theta \cos \theta + \frac{\cos \theta}{2} (\cos^2 \theta - \sin^2 \theta) \right]$$ I'm sure you can conclude from here