# MA 1505 Tutorial 11: Surface Integral, Divergence Theorem and Stokes’ Theorem

Surface Integrals of Scalar Fields: Assume $f: U \subseteq \mathbb{R}^{3} \rightarrow \mathbb{R}$ is a function, $r: D\subseteq \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$ is a surface S. Then the surface integral is $\iint_{S} f dS= \iint_{D} f(\textbf{r}(x,y)) || \textbf{r}_{x} \times \textbf{r}_{y} || dxdy$

where the left hand side is the surface integral of the scalar field and the right hand side is the multiple integration. $\textbf{r}_{x} \times \textbf{r}_{y}$ denotes the cross product between $\textbf{r}_{x}$ and $\textbf{r}_{y}$, $|| \textbf{r}_{x} \times \textbf{r}_{y} ||$ denotes the length of the vector $\textbf{r}_{x} \times \textbf{r}_{y}.$

Remark.  If $f(x,y,z)=1$ for all $(x,y,z) \in \mathbb{R}^{3}$, and $r: D\subseteq \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$ is a surface, then

the left hand side is $\iint_{S} dS = \text{ the surface area of } S.$

the right hand side is $\iint_{D} \sqrt{1+(f_{x})^{2}+ (f_{y})^{2} } dxdy$, since $\textbf{r}(x,y)=(x,y,f(x,y)), \text{ where } (x,y) \in D,$ $\textbf{r}_{x}=(1,0,f_{x})$ and $\textbf{r}_{y}=(0,1,f_{y}),$ the cross product $\textbf{r}_{x} \times \textbf{r}_{y}= (-f_{x}, -f_{y},1).$

That means: $\text{ the surface area of } S= \iint_{D} \sqrt{1+(f_{x})^{2}+(f_{y})^{2} }dxdy.$

Surface Integrals of Vector Fields:

Imagine that we have a fluid flowing through $S$, such that $\bold{F}(x)$ determines the velocity of the fluid at $\bold{x}$. The flux is defined as the quantity of fluid flowing through $S$ per unit time.

This illustration implies that if the vector field is tangent to $S$ at each point, then the flux is zero, because the fluid just flows in parallel to $S$, and neither in nor out. This also implies that if $\bold{F}$ does not just flow along $S$, that is, if $F$ has both a tangential and a normal component, then only the normal component contributes to the flux. Based on this reasoning, to find the flux, we need to take the dot product of $\bold{F}$ with the unit normal vector to $S$ at each point, which will give us a scalar field, and integrate the obtained field as above. Assume $\textbf{F} : U \subseteq \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ is a vector field, $r: D\subseteq \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$ is a surface S. Then the surface integrals of the vector field F is $\iint_{S} \textbf{F} \cdot d \textbf{S} = \iint_{S} \textbf{F} \cdot \textbf{n} dS$

The left hand side is the surface integral of vector field and the right hand side is the surface integral of scalar function, since $\textbf{F} \cdot \textbf{n}$ is a scalar function. That means, $\iint_{S} \textbf{F} \cdot d \textbf{S} = \iint_{S} \textbf{F} \cdot \textbf{n} dS = \iint_{D} \textbf{F}( \textbf{r}(x,y)) \cdot ( \textbf{r}_{x} \times \textbf{r}_{y}) dxdy$

Divergence Theorem (Gauss’s theorem or Ostrogradsky’s theorem)

This theorem is a result that relates the flow (that is, flux) of a vector field through a surface to the behavior of the vector field inside the surface. More precisely, the divergence theorem states that the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence over the region inside the surface. Intuitively, it states that the sum of all sources minus the sum of all sinks gives the net flow out of a region. $\iint_{S} \textbf{F} \cdot d \textbf{S} = \iiint_{V} \nabla \cdot \textbf{F} dV = \iiint_{V} (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}) dxdydz$

where $V \subseteq \mathbb{R}^{3}$ is a bounded domain and $\partial V=S, \textbf{F}=(P,Q,R)$ is a vector field. Stokes’ Theorem $\int_{\partial \Sigma} \textbf{F} \cdot d\textbf{r} = \iint_{\Sigma} ( \textbf{curl F} ) \cdot d \textbf{S}$

where $\textbf{curl F}= (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}) \textbf{i} + (\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}) \textbf{j} + ( \frac{\partial Q}{\partial x}- \frac{\partial P}{\partial y}) \textbf{k}$ is a vector field. $\Sigma$ is a compact surface and $\partial \Sigma$  is the boundary of $\Sigma.$ The curve $\partial\Sigma$ has the positive orientation, that means following the right hand rule. 