Force on a uniformly charged solid sphere

[stunner5000pt]

Homework Statement

Griffiths Example 8.2
Determine the net force on the northern hemisphere of a uniformly charged solid sphere of radius R and charge Q. use maxwell's stress tensor to calculate the force. (Not Gauss/Coulomb's law)

Homework Equations

The stress tensor is defined by
T_{ij} = \epsilon_{0} (E_{i}E_{j} - \frac{1}{2} \delta_{ij} E^2) + \frac{1}{\mu_{0}} (B_{i}B_{j} - \frac{1}{2} \delta_{IJ} B^2)
where i and j are coordinates

The Attempt at a Solution

ok For the bowl part
da = R^2 \sin \theta d\theta d\phi \hat{r}

\vec{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{R^2} \hat{r}

so
T_{zx} = \epsilon_{0} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin \theta \cos\theta \cos\phi

T_{zy} = \epsilon_{0} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin \theta \cos\theta \sin\phi

T_{zz} = \frac{\epsilon_{0}}{2} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 (\cos^2 \theta - \sin^2 \theta)

The net froce is in the Z direction (taking the Z axis to point stragiht up, the flat part of this bowl lying on the XY plane...

(T da)_{z} = T_{zx}da_{x}+T_{zy}da_{y}+T_{zz}da_{z} = \frac{\epsilon_{0}}{2} \left(\frac{Q}{4 \pi\epsilon_{0} R^2}\right)^2 \sin\theta \cos\theta d\theta d\phi

I don't understand what happened in that last step...
... i know this question seems silly... i should know all of this thoroughly in and out at this point...

anyway why does it turn into that??

what are dax and day and daz??
did they convert into spherical coords and then 'expand' ??

thanks for your help

 

[dextercioby]

The projections of the d\vec{a} vector onto the 3 cartesian axis.

He made all multiplications and additions and ended up with that result.

Daniel.

 

[stunner5000pt]

The projections of the d\vec{a} vector onto the 3 cartesian axis.

He made all multiplications and additions and ended up with that result.

Daniel.

i didnt see it till i actually looks at da and then i saw it

thanks for your help