How is this integral: $$\iint_{S} p(x,y,z)\,dy\,dz + q(x,y,z)\,dz\,dx + r(x,y,z)\,dx\,dy$$ equivalent to this one $$ \iint_S \vec{F}\cdot d\vec{S}$$ where $\vec{F}=\left<p,q,r\right>$
(I was trying to solve a question on Divergence theorem and spent too much time trying to find what is $\vec{F}$ and then peeked into the solution to find the above which was right in front of me the whole time)
Thank you in advance
The second version is what I call the “20th-century” notation for the surface integral. The definition is: $$ \iint_S \vec F \cdot d \vec S = \iint_S \vec F \cdot \vec n \,dS $$ where $\vec n$ is a unit normal vector field on $S$.
To compute this integral, we normally need a parametrization. Suppose that $D$ is a domain of $\mathbb R^2$, and $\vec r \colon D \to S$ is a parametrization of $S$ compatible with the orientation $\vec n$. Write $\vec r(u,v) = (x(u,v),y(u,v),z(u,v))$. We define \begin{align*} \vec r_u &= \left<\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}\right> \\ \vec r_v &= \left<\frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v}\right> \end{align*} Then $\vec r$ is compatible with $\vec n$ if $\vec r_u \times \vec r_v$ is in the positive direction of $\vec n$ for all $(u,v) \in D$. In this case, $$ \iint_S \vec F \cdot \vec n \,dS = \iint_D \vec F(\vec r(u,v)) \cdot (\vec r_u \times \vec r_v)\,dA $$ Let's take it one step further down. Write $\vec F(x,y,z) = \left<P(x,y,z),Q(x,y,z),R(x,y,z)\right>$. Then $$ \vec r_u \times \vec r_v = \left<\frac{\partial y}{\partial u} \frac{\partial z}{\partial v} -\frac{\partial z}{\partial u} \frac{\partial y}{\partial v}, \frac{\partial z}{\partial u} \frac{\partial x}{\partial v} -\frac{\partial x}{\partial u} \frac{\partial z}{\partial v}, \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} -\frac{\partial y}{\partial u} \frac{\partial x}{\partial v} \right> $$ So \begin{align*} \iint_D \vec F(\vec r(u,v)) \cdot (\vec r_u \times \vec r_v)\,dA &= \iint_D \biggl( P(x(u,v),y(u,v),z(u,v)) \left( \frac{\partial y}{\partial u} \frac{\partial z}{\partial v} -\frac{\partial z}{\partial u} \frac{\partial y}{\partial v} \right) \\&\qquad\qquad+ Q(x(u,v),y(u,v),z(u,v)) \left( \frac{\partial z}{\partial u} \frac{\partial x}{\partial v} -\frac{\partial x}{\partial u} \frac{\partial z}{\partial v} \right) \\&\qquad\qquad+ R(x(u,v),y(u,v),z(u,v)) \left( \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} -\frac{\partial y}{\partial u} \frac{\partial x}{\partial v} \right) \biggr)\,du\,dv\tag{$*$} \end{align*}
In order to understand the “19th-century” notation: $$ \iint_S \left(P \,dy\,dz + Q \,dz\,dx + R \,dx\,dy\right) $$ you just have to treat the $dx$, $dy$, $dz$ symbols as differentials, compute them in terms of $du$ and $dv$, and assume that differentials can be multiplied skew-commutatively. So: \begin{align*} dx &= \frac{\partial x}{\partial u}\,du + \frac{\partial x}{\partial v}\,dv \\ dy &= \frac{\partial y}{\partial u}\,du + \frac{\partial y}{\partial v}\,dv \\ dz &= \frac{\partial z}{\partial u}\,du + \frac{\partial z}{\partial v}\,dv \end{align*} and therefore, \begin{align*} dy\,dz &= \left(\frac{\partial y}{\partial u}\,du + \frac{\partial y}{\partial v}\,dv\right) \left(\frac{\partial z}{\partial u}\,du + \frac{\partial z}{\partial v}\,dv\right) \\&= \frac{\partial y}{\partial u}\frac{\partial z}{\partial u} \,du\,du + \frac{\partial y}{\partial u} \frac{\partial z}{\partial v}\,du\,dv + \frac{\partial y}{\partial v}\frac{\partial z}{\partial u}\,dv\,du +\frac{\partial y}{\partial v}\frac{\partial z}{\partial v}\,dv\,dv \\&= \left(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial y}{\partial v}\frac{\partial z}{\partial u}\right)\,du\,dv \end{align*} (Notice $du\,du = dv\,dv = 0$, and also $dv\,du = -du\,dv$, by skew-commutativity.) You see that this is the same combination of partial derivatives that forms the $x$-component of $\vec r_u \times \vec r_v$. Similarly, \begin{align*} dz\,dx &= \left( \frac{\partial z}{\partial u} \frac{\partial x}{\partial v} -\frac{\partial x}{\partial u} \frac{\partial z}{\partial v} \right)\,du\,dv \\ dy\,dy &= \left( \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} -\frac{\partial y}{\partial u} \frac{\partial x}{\partial v} \right)\,du\,dv \end{align*} So $ \iint_S \left(P \,dy\,dz + Q \,dz\,dx + R \,dx\,dy\right)$ turns out to be equal to the right-hand side of ($*$) as well.