PX153 - I11 - surface integrals

\iint_{S} ϕ d S o r \iint_{S} \vec{a} \cdot d \vec{S} o r \iint_{S} \vec{a} \times d \vec{S} o r \iint_{S} a d \vec{S}

$d \vec{S}$ is a surface element that represents $d S \hat{n}$ , where $d S$ is an infinitesimal area, and $\hat{n}$ is the normal to the surface element (3D vector with a magnitude of 1)
formally, dividing $S$ into $N$ elements, each with an area, $Δ S_{p}$ , where $p = 1, . . . N$ , and each surface element is assigned its normal, ${\hat{n}}_{p}$ :

\iint_{S} \vec{a} \cdot d \vec{S} = lim_{N \to \infty} \sum_{p = 1}^{N} \vec{a} ({\bar{x}}_{p}, {\bar{y}}_{p}, {\bar{z}}_{p}) \cdot Δ S_{p} {\hat{n}}_{p}

d \vec{S} = r^{2} \sin θ d θ d ϕ

eg: what is the vector area of the hemisphere? $I = \int_{S} 1 d \vec{S}$ $x^{2} + y^{2} + z^{2} = a^{2}$
$d \vec{S} = a^{2} \sin θ d θ d ϕ \hat{r}$

\begin{aligned} I & = \int_{0}^{2 π} \int_{0}^{\frac{π}{2}} (\sin θ \cos ϕ \hat{i} + \sin θ \sin ϕ \hat{j} + \cos θ \hat{k}) a^{2} \sin θ d θ d ϕ \\ = \int_{0}^{2 π} \int_{0}^{\frac{π}{2}} \cos θ \hat{k} a^{2} \sin θ d θ d ϕ \\ = a^{2} \hat{k} \int_{0}^{2 π} \frac{1}{2} {[- \frac{\cos (2 θ)}{2}]}_{0}^{\frac{π}{2}} d ϕ \\ = a^{2} \hat{k} \int_{0}^{2 π} \frac{1}{2} d ϕ \\ = a^{2} π d ϕ \end{aligned}

flat surface

eg: $S$ is area $0 \leq x \leq 1, 0 \leq x \leq 1$ . compute: $I = \iint \vec{\nabla} ϕ \cdot d \vec{S}$ , where, $ϕ = x y z$
- convention: the sense of the normal is defined by the orientation at the boundary: follows the right hand thumb rule
  $\vec{\nabla} ϕ = (y z, x z, x y)$
  $d \vec{S} = (0, 0, d x d y)$ (integrating over an infinitesimal area, $d x d y$ )
$I = \int_{0}^{1} \int_{0}^{1} x y d x d y = . . . = \frac{1}{4}$