1 Surface integrals involving vectors

1.1 The unit normal

For the surface of any three-dimensional shape, it is possible to find a vector lying perpendicular to the surface and with magnitude 1. The unit vector points outwards from the surface and is usually denoted by $\underset{̲}{\stackrel{̂}{n}}$ .

Example 16

If $S$ is the surface of the sphere $x^2+y^2+z^2=a^2$ find the unit normal $\underset{̲}{\stackrel{̂}{n}}$ .

Solution

The unit normal at the point $P\left(x,y,z\right)$ points away from the centre of the sphere i.e. it lies in the direction of $x\underset{̲}{i}+y\underset{̲}{j}+z\underset{̲}{k}$ . To make this a unit vector it must be divided by its magnitude $\sqrt{x^2+y^2+z^2}$ i.e. the unit vector

where $a=\sqrt{x^2+y^2+z^2}$

Figure 7:

Example 17

For the cube $0\le x\le 1$ , $0\le y\le 1$ , $0\le z\le 1$ , find the unit normal $\underset{̲}{\stackrel{̂}{n}}$ for each face.

Solution

On the face given by $x=0$ , the unit normal points in the negative $x$ -direction. Hence the unit normal is $-\underset{̲}{i}$ . Similarly :-

On the face $x=1$ the unit normal is $\underset{̲}{i}$ . On the face $y=0$ the unit normal is $-\underset{̲}{j}$ .

On the face $y=1$ the unit normal is $\underset{̲}{j}$ . On the face $z=0$ the unit normal is $-\underset{̲}{k}$ .

On the face $z=1$ the unit normal is $\underset{̲}{k}$ .

1.2 $\underset{̲}{dS}$ and the unit normal

The vector $\underset{̲}{dS}$ is a vector, being an element of the surface with magnitude $dudv$ and direction perpendicular to the surface.



If the plane in question is the $Oxy$ plane, then $\underset{̲}{dS}=\underset{̲}{\stackrel{̂}{n}}dudv=\underset{̲}{k}dxdy$ .

Figure 8:

If the plane in question is not one of the three coordinate planes ( $Oxy$ , $Oxz$ , $Oyz$ ), appropriate adjustments must be made to express $\underset{̲}{dS}$ in terms of two of $dx$ and $dy$ and $dz$ .

Example 18

The rectangle $OABC$ lies in the plane $z=y$ (Figure 9).

The vertices are $O=\left(0,0,0\right)$ , $A=\left(1,0,0\right)$ , $B=\left(1,1,1\right)$ and $C=\left(0,1,1\right)$ .

Find a unit vector $\underset{̲}{\stackrel{̂}{n}}$ normal to the plane and an appropriate vector $\underset{̲}{dS}$ expressed in terms of $dx$ and $dy$ .

Figure 9:

Solution

Note that two vectors in the rectangle are $\overrightarrow{OA}=\underset{̲}{i}$ and $\overrightarrow{OC}=\underset{̲}{j}+\underset{̲}{k}$ . A vector perpendicular to the plane is $\underset{̲}{i}\times \left(\underset{̲}{j}+\underset{̲}{k}\right)=-\underset{̲}{j}+\underset{̲}{k}$ . However, this vector is of magnitude $\sqrt{2}$ so the unit normal vector is $\underset{̲}{\stackrel{̂}{n}}=\frac{1}{\sqrt{2}}\left(-\underset{̲}{j}+\underset{̲}{k}\right)=-\frac{1}{\sqrt{2}}\underset{̲}{j}+\frac{1}{\sqrt{2}}\underset{̲}{k}$ .

The vector $\underset{̲}{dS}$ is therefore $\left(-\frac{1}{\sqrt{2}}\underset{̲}{j}+\frac{1}{\sqrt{2}}\underset{̲}{k}\right)dudv$ where $du$ and $dv$ are increments in the plane of the rectangle $OABC$ . Now, one increment, say $du$ , may point in the $x$ -direction while $dv$ will point in a direction up the plane, parallel to $OC$ . Thus $du=dx$ and (by Pythagoras) $dv=\sqrt{{\left(dy\right)}^2+{\left(dz\right)}^2}$ . However, as $z=y$ , $dz=dy$ and hence $dv=\sqrt{2}dy$ .

Thus, $\underset{̲}{dS}=\left(-\frac{1}{\sqrt{2}}\underset{̲}{j}+\frac{1}{\sqrt{2}}\underset{̲}{k}\right)dx\sqrt{2}dy=\left(-\underset{̲}{j}+\underset{̲}{k}\right)dxdy$ .

Note :- the factor of $\sqrt{2}$ could also have been found by comparing the area of rectangle $OABC$ , i.e. $1$ , with the area of its projection in the $Oxy$ plane i.e. $OADE$ with area $\frac{1}{\sqrt{2}}$ .

1.3 Integrating a scalar field

A function can be integrated over a surface by constructing a double integral and integrating in a manner similar to that shown in HELM booklet  27.1 and HELM booklet  27.2. Often, such integrals can be carried out with respect to an element containing the unit normal.

Example 19

Evaluate the integral

over the area $A$ where $A$ is the square $0\le x\le 1$ , $0\le y\le 1$ , $z=0$ .

Solution

In this integral, $\underset{̲}{dS}$ becomes $\underset{̲}{k}dxdy$ i.e. the unit normal times the surface element. Thus the integral is

Example 20

Find $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S}u\underset{̲}{dS}$ where $u=r^2=x^2+y^2+z^2$ and $S$ is the surface of the unit cube $0\le x\le 1$ , $0\le y\le 1$ , $0\le z\le 1$ .

Solution

The unit cube has six faces and the unit normal vector $\underset{̲}{\stackrel{̂}{n}}$ points in a different direction on each face. The surface integral must be evaluated for each face separately and the results summed.

On the face $x=0$ , the unit normal $\underset{̲}{\stackrel{̂}{n}}=-\underset{̲}{i}$ and the surface integral is

On the face $x=1$ , the unit normal $\underset{̲}{\stackrel{̂}{n}}=\underset{̲}{i}$ and the surface integral is

The net contribution from the faces $x=0$ and $x=1$ is $-\frac{2}{3}\underset{̲}{i}+\frac{5}{3}\underset{̲}{i}=\underset{̲}{i}$ .

Due to the symmetry of the scalar field $u$ and the unit cube, the net contribution from the faces $y=0$ and $y=1$ is $\underset{̲}{j}$ while the net contribution from the faces $z=0$ and $z=1$ is $\underset{̲}{k}$ .

The sum i.e. the surface integral $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S}u\underset{̲}{dS}=\underset{̲}{i}+\underset{̲}{j}+\underset{̲}{k}$

When the surface does not lie in one of the planes $Oxy$ , $Oxz$ , $Oyz$ , extra care must be taken when finding $\underset{̲}{dS}$ .

Example 21

Find $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S}f\underset{̲}{dS}$ where $f$ is the function $2x$ and $S$ is the surface of the triangle bounded by $\left(0,0,0\right)$ , $\left(0,1,1\right)$ and $\left(1,0,1\right)$ . (See Figure 10.)

Figure 10:

Solution

The unit vector $\underset{̲}{n}$ is perpendicular to two vectors in the plane e.g. $\left(\underset{̲}{j}+\underset{̲}{k}\right)$ and $\left(\underset{̲}{i}+\underset{̲}{k}\right)$ . The vector $\left(\underset{̲}{j}+\underset{̲}{k}\right)\times \left(\underset{̲}{i}+\underset{̲}{k}\right)=\underset{̲}{i}+\underset{̲}{j}-\underset{̲}{k}$ and has magnitude $\sqrt{3}$ . Hence the normal vector $\stackrel{̂}{\underset{̲}{n}}=\frac{1}{\sqrt{3}}\underset{̲}{i}+\frac{1}{\sqrt{3}}\underset{̲}{j}-\frac{1}{\sqrt{3}}\underset{̲}{k}$ .

As the area of the triangle is $\frac{\sqrt{3}}{2}$ and the area of its projection in the $Oxy$ plane is $\frac{1}{2}$ , the vector $\underset{̲}{dS}=\frac{\sqrt{3}∕2}{1∕2}\stackrel{̂}{\underset{̲}{n}}dydx=\left(\underset{̲}{i}+\underset{̲}{j}+\underset{̲}{k}\right)dydx$ .

Thus

The scalar function being integrated may be the divergence of a suitable vector function.

Example 22

Find $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\cdot \underset{̲}{F}\right)\underset{̲}{dS}$ where $\underset{̲}{F}=2x\underset{̲}{i}+yz\underset{̲}{j}+xy\underset{̲}{k}$ and $S$ is the surface of the triangle with vertices at $\left(0,0,0\right)$ , $\left(1,0,0\right)$ and $\left(1,1,0\right)$ .

Solution

Note that $\underset{̲}{\nabla }\cdot \underset{̲}{F}=2+z=2$ as $z=0$ everywhere along $S$ . As the triangle lies in the $Oxy$ plane, the normal vector $\underset{̲}{n}=\underset{̲}{k}$ and $\underset{̲}{dS}=\underset{̲}{k}dydx$ .

Thus,

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\cdot \underset{̲}{F}\right)\underset{̲}{dS}={\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{y=0}^{x}2dydx\underset{̲}{k}={\int \; <!--nolimits\; -->}_0^1{\left[2y\right]}_0^{x}dx\underset{̲}{k}={\int \; <!--nolimits\; -->}_0^{1}2xdx\underset{̲}{k}={\left[x^2\right]}_0^1\underset{̲}{k}=\underset{̲}{k}$

Task!

Evaluate the integral $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S}4x\underset{̲}{dS}$ where $S$ represents the trapezium with vertices at $\left(0,0\right)$ , $\left(3,0\right)$ , $\left(2,1\right)$ and $\left(0,1\right)$ .

1. Find the vector $\underset{̲}{dS}$ :

   Answer

   $\underset{̲}{k}dxdy$

2. Write the surface integral as a double integral:

   Answer

   ${\int \; <!--nolimits\; -->}_{y=0}^1{\int \; <!--nolimits\; -->}_{x=0}^{3-y}4xdxdy\underset{̲}{k}$ .  The range of values of $y$ is $y=0$ to $y=1$ .

     For each value of $y$ , $x$ varies from $x=0$ to $x=3-y$

3. Evaluate this double integral:

   Answer

   $\frac{38}{3}\underset{̲}{k}$

Exercises

1. Evaluate the integral $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S}xy\underset{̲}{dS}$ where $S$ is the triangle with vertices at $\left(0,0,4\right)$ , $\left(0,2,0\right)$ and $\left(1,0,0\right)$ .
2. Find the integral $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{S}xyz\underset{̲}{dS}$ where $S$ is the surface of the unit cube $0\le x\le 1$ , $0\le y\le 1$ , $0\le z\le 1$ .
3. Evaluate the integral $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left[\underset{̲}{\nabla }\cdot \left(x^2\underset{̲}{i}+yz\underset{̲}{j}+x^{2}y\underset{̲}{k}\right)\right]\underset{̲}{dS}$ where $S$ is the rectangle with vertices at $\left(1,0,0\right)$ , $\left(1,1,0\right)$ , $\left(1,1,1\right)$ and $\left(1,0,1\right)$ .

Answer

1.4 Integrating a vector field

In a similar manner to the case of a scalar field, a vector field may be integrated over a surface. Two common integrals are ${\int \; <!--nolimits\; -->}_S\underset{̲}{F}\left(\underset{̲}{r}\right)\cdot \underset{̲}{dS}$ and ${\int \; <!--nolimits\; -->}_S\underset{̲}{F}\left(\underset{̲}{r}\right)\times \underset{̲}{dS}$ which integrate to a scalar and a vector respectively. Again, when $\underset{̲}{dS}$ is expressed appropriately, the expression will reduce to a double integral. The form ${\int \; <!--nolimits\; -->}_S\underset{̲}{F}\left(r\right)\underset{̲}{dS}$ has many important applications, e.g. flux.

Example 23

Evaluate the integral

over the area $A$ where $A$ is the square $0\le x\le 1$ , $0\le y\le 1$ , $z=0$ .

Solution

On $A$ , the unit normal is $dxdy\underset{̲}{k}$ so the integral becomes

Example 24

Evaluate ${\int \; <!--nolimits\; -->}_A\underset{̲}{r}\cdot \underset{̲}{dS}$ where $A$ represents the surface of the unit cube $0\le x\le 1$ , $0\le y\le 1$ , $0\le z\le 1$ and $\underset{̲}{r}$ represents the vector $x\underset{̲}{i}+y\underset{̲}{j}+z\underset{̲}{k}$ .

Solution

The vector $\underset{̲}{dS}$ (in the direction of the normal vector) will be a constant vector on each face, but will be different for each face.

On the face $x=0$ (left), $\underset{̲}{dS}=-dydz\underset{̲}{i}$ and the integral on this face is

${\int \; <!--nolimits\; -->}_{z=0}^1{\int \; <!--nolimits\; -->}_{y=0}^1\left(0\underset{̲}{i}+y\underset{̲}{j}+z\underset{̲}{k}\right)\cdot \left(-dydz\underset{̲}{i}\right)={\int \; <!--nolimits\; -->}_{z=0}^1{\int \; <!--nolimits\; -->}_{y=0}^{1}0dydz=0$

Similarly on the face $y=0$ (front), $\underset{̲}{dS}=-dxdz\underset{̲}{j}$ and the integral on this face is

${\int \; <!--nolimits\; -->}_{z=0}^1{\int \; <!--nolimits\; -->}_{x=0}^1\left(x\underset{̲}{i}+0\underset{̲}{j}+z\underset{̲}{k}\right)\cdot \left(-dxdz\underset{̲}{j}\right)={\int \; <!--nolimits\; -->}_{z=0}^1{\int \; <!--nolimits\; -->}_{x=0}^{1}0dxdz=0$

Furthermore on the face $z=0$ (bottom), $\underset{̲}{dS}=-dxdy\underset{̲}{k}$ and the integral on this face is

${\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{y=0}^1\left(x\underset{̲}{i}+y\underset{̲}{j}+0\underset{̲}{k}\right)\cdot \left(-dxdy\underset{̲}{k}\right)={\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{y=0}^{1}0dxdy=0$

On these three faces, the contribution to the integral is zero. However, on the face $x=1$ (right), $\underset{̲}{dS}=+dydz\underset{̲}{i}$ and the integral on this face is

${\int \; <!--nolimits\; -->}_{z=0}^1{\int \; <!--nolimits\; -->}_{y=0}^1\left(1\underset{̲}{i}+y\underset{̲}{j}+z\underset{̲}{k}\right)\cdot \left(+dydz\underset{̲}{i}\right)={\int \; <!--nolimits\; -->}_{z=0}^1{\int \; <!--nolimits\; -->}_{y=0}^{1}1dydz=1$

Similarly, on the face $y=1$ (back), $\underset{̲}{dS}=+dxdz\underset{̲}{j}$ and the integral on this face is

${\int \; <!--nolimits\; -->}_{z=0}^1{\int \; <!--nolimits\; -->}_{x=0}^1\left(x\underset{̲}{i}+1\underset{̲}{j}+z\underset{̲}{k}\right)\cdot \left(+dxdz\underset{̲}{j}\right)={\int \; <!--nolimits\; -->}_{z=0}^1{\int \; <!--nolimits\; -->}_{x=0}^{1}1dxdz=1$

and finally,on the face $z=1$ (top), $\underset{̲}{dS}=+dxdy\underset{̲}{k}$ and the integral on this face is

${\int \; <!--nolimits\; -->}_{y=0}^1{\int \; <!--nolimits\; -->}_{x=0}^1\left(x\underset{̲}{i}+y\underset{̲}{j}+1\underset{̲}{k}\right)\cdot \left(+dxdy\underset{̲}{k}\right)={\int \; <!--nolimits\; -->}_{y=0}^1{\int \; <!--nolimits\; -->}_{x=0}^{1}1dxdy=1$

Adding together the contributions from the various faces gives ${\int \; <!--nolimits\; -->}_A\underset{̲}{r}\cdot \underset{̲}{dS}=0+0+0+1+1+1=3$