# Math2111: Chapter 4: Surface integrals. Section 3: Surface integrals of vector fields

In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

In Section 2 we have seen how to calculate surface area and integrate scalar fields over surfaces. We now establish an extension similar to the step from scalar line integrals to vector line integrals.

We assume again that the surfaces we consider are smooth, that is, they are assumed to be images of parameterised surfaces $\boldsymbol{\Phi}:D\to\mathbb{R}^3$ for which:

• $D$ is a non-empty, compact and Jordan-measurable subset of $\mathbb{R}^2$;
• the mapping $\boldsymbol{\Phi}$ is one-to-one;
• $\boldsymbol{\Phi}$ is continuously differentiable;
• the normal vector $\boldsymbol{n}=\frac{\partial \boldsymbol{\Phi}}{\partial u} \times \frac{\partial \boldsymbol{\Phi}}{\partial v} \neq \boldsymbol{0}$ except possibly at a finite number of points;

(Notice, the condition that ${}D$ is compact can also be replaced by the condition that the surface $S=\{\boldsymbol{\Phi}(u,v): (u,v)\in D\}$ is compact.)

Definition
Let $\boldsymbol{F}:\mathbb{R}^3\to\mathbb{R}^3$ be a vector field and $S$ be the image of a parameterised surface $\boldsymbol{\Phi}:D\subset\mathbb{R}^2\to\mathbb{R}^3.$ Then we define the surface integral by

$\iint_{S} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{\mathcal{S}} = \iint_D \boldsymbol{F}(\boldsymbol{\Phi}(u,v))\cdot \left(\frac{\partial \boldsymbol{\Phi}}{\partial u}(u,v) \times \frac{\partial \boldsymbol{\Phi}}{\partial v}(u,v) \right) \mathrm{d}u \,\mathrm{d} v.$

Notice that we have

$\displaystyle \iint_S \boldsymbol{F} \cdot\mathrm{d}\boldsymbol{\mathcal{S}} = \iint_D \boldsymbol{F} \cdot \boldsymbol{n} \,\mathrm{d} \mathcal{S},$

where $\boldsymbol{n}$ denotes the normal vector to the surface $S$. Hence the surface integral extends the line integral in normal form by increasing the dimension by one. Further, also the physical interpretation of line integrals in normal form can be extended to this case. Therefore the surface integral of a vector field is also called the flux of the vector field across the surface.

Let $\boldsymbol{F}=(F_1,F_2,F_3)$, then the surface integral can also be written as

$\displaystyle \begin{array}{rcl} \iint_S \boldsymbol{F}\cdot \mathrm{d}\boldsymbol{\mathcal{S}} &=& \iint_D \left(F_1 \frac{\partial (Y,Z)}{\partial (u,v)} + F_2 \frac{\partial (Z,X)}{\partial (u,v)} + F_3 \frac{\partial (X,Y)}{\partial (u,v)}\right)\,\mathrm{d}u \,\mathrm{d} v \\ && \\ &=& \iint_D F_1 \,\mathrm{d} (Y,Z) + F_2 \,\mathrm{d} (Z,X) + F_3 \,\mathrm{d} (X,Y).\end{array}$

Orientation of surfaces

Recall that for line integrals in normal form, one has to choose one of the two possible choices for a normal vector. For surface integral one also needs to define which normal vector should be used. Notice that we assume that the parameterisation is smooth and that the normal vector $\boldsymbol{n}\neq \boldsymbol{0}$ except for a finite number of points. Hence the normal vector is a continuous vector field defined on the surface.

The situation is more complicated though, since there are surfaces which do not have a normal vector which is well-defined. Note that if we choose a normal vector on a surface, then the surface has two sides: one in which the normal vector $\boldsymbol{n}$ point and one in which the normal vector $-\boldsymbol{n}$ points. But this is not always possible. The most famous example of such a surface is the Möbius strip. For this surface one cannot define a continuous normal vector which is nonzero except for a finite number of points.

The Möbius strip can for instance be generated by setting

$\displaystyle \boldsymbol{\Phi}(u,v) = X(u,v)\widehat{\boldsymbol{i}} + Y(u,v) \widehat{\boldsymbol{j}} + Z(u,v) \widehat{\boldsymbol{k}},$

where

$\displaystyle \begin{array}{rcl} X(u,v) &=& \frac{v}{2} \sin \frac{u}{2}, \\ && \\ Y(u,v) &=& \left(1+\frac{v}{2}\cos\frac{u}{2}\right)\sin u, \\ &&\\ Z(u,v)&=& \left(1+\frac{v}{2}\cos\frac{u}{2}\right)\cos u, \end{array}$

where $0 \le u\le 2\pi$ and $- 1 \le v \le 1.$

The following figure shows the Möbius strip.

[The Mobius strip can be generated in Matlab using the following commands:
>> u=linspace(0,2*pi,30);
>> v=linspace(-1,1,15);[u,v]=meshgrid(u,v);
>> u=linspace(0,2*pi,30);
>> v=linspace(-1,1,15);
>> [u,v]=meshgrid(u,v);
>> z=(1+v/2.*cos(u/2)).*cos(u);
>> y=(1+v/2.*cos(u/2)).*sin(u);
>> x=v/2.*sin(u/2);
>> surf(x,y,z)
]

Example
Calculate the flux of $\boldsymbol{F}(x,y,z)=(x,y,1)$ across the paraboloid

$\displaystyle z= 4-(x^2+y^2),$

where $z \ge 0,$ oriented with upward pointing normal vector.

We set $g(x,y)=z=4-x^2-y^2$ and define the parameterisation $\boldsymbol{\Phi}(x,y)=(x,y,g(x,y))$. Hence we obtain the normal vector by

$\displaystyle \boldsymbol{n}= \left(-\frac{\partial g}{\partial x}, -\frac{\partial g}{\partial y}, 1\right) =(2x,2y,1).$

This vector points upwards since the last component is bigger than zero. From $z \ge 0$ we obtain $0\le x^2+y^2\le 4$. Therefore we have

$\displaystyle \iint_S \boldsymbol{F}\cdot \boldsymbol{n} \,\mathrm{d} A= \iint_D (x,y,1)\cdot (2x,2y,1) \,\mathrm{d} x\,\mathrm{d} y = \int_0^2 \int_0^{2\pi} (2r^2+1) r \,\mathrm{d} \theta\,\mathrm{d} r,$

where in the last step we switched to polar coordinates. Hence we obtain $\iint_S \boldsymbol{F}\cdot \boldsymbol{\mathcal{S}}=20\pi.$
$\Box$

Exercise
Let the surface $S$ be given as the graph of a function $z = g(x,y)$ defined on a domain $D\subset \mathbb{R}^2$ (that is, the surface is $S= \{(x,y,(g(x,y)): (x,y)\in D\}$). Let the vector field $\boldsymbol{F}= \widehat{\boldsymbol{k}}.$

• Show that $\iint_S \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{\mathcal{S}}= \mbox{area }(D).$
• Give a geometrical interpretation of this result.

$\Box$