Calculating equation of motion from action Suppose my action integral is $$S=\int d^4x(\nabla \times A)^2$$ and $\delta S$ gives $$\delta S =\int d^4x [2(\nabla \times A).(\nabla \times \delta A)]$$
I would like to calculate the coefficient of $\delta A$ from this action integral. But I am stuck. How can I separate the $\delta A$ from the term like this?
 A: Let's do what  Heidar says and write it with indices, and identify the Lagrangian.
$$
L=\frac{1}{2}(\vec{\nabla}\times \vec{A})^2 = \frac{1}{2}\epsilon_{ijk}\partial_j A_k \epsilon_{ilm}\partial_l A_m
$$
where, if you haven't heard of it yet, you pretend there is a summation symbol for each repeated index.  Then since there are no bare $A_i$ sitting by themselves, only $\partial_i A_j$s the only part of Lagrange's equations that will contribute are
$$
\partial_q \frac{\partial L}{\partial (\partial_q A_p)}
$$
which we set equal to zero following the equations.  Then
$$
\frac{\partial L}{\partial (\partial_q A_p)}=\frac{1}{2}(\epsilon_{ijk}\delta_{jq}\delta_{kp}\epsilon_{ilm}\partial_l A_m+\epsilon_{ijk}\partial_j A_k \epsilon_{ilm}\delta_{lq}\delta_{mp})
$$
using
$$
\frac{\partial (\partial_i A_j)}{\partial (\partial_q A_p)}=\delta_{iq}\delta_{jp}.
$$
Then we have
$$
\partial_q \frac{\partial L}{\partial (\partial_q A_p)}=\partial_q(\epsilon_{iqp}\epsilon_{ijk}\partial_i A_j) = \partial_q ((\delta_{qj}\delta_{pk}-\delta_{qk}\delta_{pj})\partial_{i} A_j)=\partial_q (\partial_q A_p - \partial_p A_q)=0
$$
where i used the contracted epsilon identity and changed the repeated indices as i needed them in order to combine terms.  Hope this helps.
EDIT:
Well, I'll still try and help out, hopefully I don't make anything any worse.
$$
\delta S = \int d^3 x \frac{1}{2}\epsilon_{ijk}\epsilon_{ilm}\delta(\partial_j A_k)\partial_l A_m+\int d^3 x\frac{1}{2}\epsilon_{ijk}\epsilon_{ilm}\partial_j A_k \delta(\partial_l A_m)
$$
Now with the variations $\delta$ we can interchange the order of $\partial$ and $\delta$
$$
\delta(\partial_i A_j)=\partial_i (\delta A_k)
$$
So with the two terms multiplied above we get
$$
\partial_j( \delta A_k)\partial_l A_m=\partial_j (\delta A_k \partial_l A_m)-\delta A_k \partial_j \partial_l A_m
$$
from the product rule.  This helps isolate the variation of the field.  Please (everyone) let me know if this is still confusing and/or wrong.  Hope this helps.
