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It is, in fact, a double integral! The first notation used $$\varPhi_E = \oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \oint_S \vec{E} \cdot \hat{n} \ \mathrm{d}A$$ is simply a more compact notation. It's much easier to write $\mathrm{d} \vec{A}$ instead of, say, $r \ \mathrm{d}r \ \mathrm{d}\theta$ all the time. Furthermore, it's more general, as $\mathrm{d} ... 3 The general formula is indeed a double integral, so the most technically correct way to write it is $$\Phi_E = \iint_S \vec{E}\cdot\mathrm{d}^2\vec{A}$$ But when formulas start to involve four, five, or more integrals, it gets tedious to write them all out all the time, so there's a notational convention in which a multiple integration can be designated by ... 3 It is definitely an ambiguous notation, but one that is quite conventional. You should interpret it as:$(\partial_a X_\mu)(\partial^a X^\mu)$. For instance, often the Klein-Gordon Lagrangian is written as: $$\mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi + \cdots$$ which should be interpreted as: $$\mathcal{L} = \frac{1}{2} (\partial_\mu ... 2 The Kronecker delta is used in the first term, not the second. In the first term, replace \sum\limits_i \omega_i^2 with \sum\limits_{i,j}\omega_i\omega_j\delta_{ij}. For the second term, rearranging the summations is a pretty common thing to do. The identity that you are interested in is essentially the distributive property. As a simple example,$$ ... 1 All the other answers that say the "single" integral is simply a shorthand notation are right, but it is well to remember that one can indeed construe the integral as a single integral as a Lebesgue integral (if you do nothing else, look up Lebesgue's very cute little half paragraph summary (on the Wiki page) of his idea in a letter to Paul Montel). If ... 1 There is an official convention for a positive muon being a nucleus, that a positive muon with one electron is muonium (Mu) and a positive muon with two electrons is muonide (Mu-). See Names for Muonic and Hydrogen Atoms and Their Ions. For a negative muon replacing an electron in helium, I see both$He\mu$and$^{4.1}H\$ in the same paper: Kinetic Isotope ...