Surface integrals

Scalar surface integrals

Let us first consider a simple case of evaluating surface integrals. Suppose we have a surface, $\mathcal{S}$ given by $z=g(x, y)$. Then, we define the scalar surface integral of a function $f$ [where $f=f(x, y, z)$ ] over a surface $\mathcal{S}$ (note that this is analogous to Eq. (2.34) ) as follows:

where $d S$ is the surface differential.

Definition 2.8 The surface integral in which the surface $\mathcal{S}$ is given by $z=g(x, y)$ is computed from:

Note that the integral on the RHS of Eq. (2.98) is a double integral of the form encountered in the previous section in these notes. Note also that by setting $z=g(x, y)$, we choose a surface that lies above some region $\mathcal{R}$ in the $x y$-plane. We can also consider surface integrals for a surface $y=g(x, z)$ that lies above a region $\mathcal{R}$ in the $x z$-plane or a surface $x=g(y, z)$ that lies above a region in the $y z$-plane.

Now, let us consider more complicated examples which will require parametrisation. We saw that the parametrisation of curves is key when considering line integrals; in the case of surface integrals, we need to discuss the the parametrisation of surfaces. An example surface is described by:

In Eq. (2.99), the variables $u$ and $v$ are the parameters (analogous to $t$ in the case of line integrals), defined in a domain $D$ known as the parameter domain. Since the surface is two-dimensional, we now need two parameters to define it. For example, a surface $\mathcal{S}$ is given by the following equation:

where the surface consists of all points $(x, y, z)$ such that: $x=u+v, y=$ $u^{3}-v, z=v^{3}-u$, for $(u, v)$ in a two-dimensional domain, $D$. In what follows, we always assume that $\boldsymbol{G}$ is continuously differentiable: we can always compute the partial derivatives of the functions $x(u, v), y(u, v)$ and $z(u, v)$. Suppose we have a curve that is given by the parametrisation given by Eq. (2.99). In this case, the scalar surface integral is given by Theorem 2.8 .

Computing a scalar surface integral

Let $\boldsymbol{G}(u, v)$ be a parametrisation for a surface $\mathcal{S}$ in some region $\mathcal{R}$, where $\mathcal{R}$ is the range of the parameters (i.e. $u$ and $v$ ) that trace the surface $\mathcal{S}$ in the chosen $u v$ plane. If $f$ and $\boldsymbol{G}, \frac{\partial \boldsymbol{G}}{\partial u}$ and $\frac{\partial \boldsymbol{G}}{\partial v}$ are continuous, then:

where $\frac{\partial \boldsymbol{G}}{\partial u} \times \frac{\partial \boldsymbol{G}}{\partial v}$ is the cross product of the partial derivatives of $\boldsymbol{G}$ and the partial derivatives are:

In Eq. (2.101), the absolute value given by

is called the area element and it represents a small area $d S$ of the surface obtained by changing the coordinates $u$ and $v$ by small amounts $d u$ and $d v$. Let us look at an example next.

Just like we talked about oriented curves in Subsec. 2.3.2 here, we introduce the concept of oriented surfaces. At each point on the surface, we have a unit normal vector, $\boldsymbol{n}$. We can have two normal directions at each point on the surface, $\mathcal{S}$; for example, in the case of a sphere, the unit normal vector can be pointing to the outside or the inside of the sphere. If $\boldsymbol{n}$ is the unit normal vector determined by one orientation then, $-\boldsymbol{n}$ is the unit normal vector determined by the opposite orientation. Example 2.11 Evaluate the following integral:

where $\mathcal{S}$ is part of a cylindrical surface parametrised by:

where $0 \leq u \leq 2 \pi$ and $0 \leq v \leq H$, and $H$ is a constant height.

Solution We need to use Eq. (2.101) to evaluate the surface integral. Since we are given the parametrisation, we compute the partial derivatives of $\boldsymbol{G}$ as follows:

The cross product is:

It follows that the area element is calculated as:

Now we can evaluate the surface integral:

The integral above yields $\mathcal{I}=2 \pi a \ln \left(\frac{H+\sqrt{a^{2}+H^{2}}}{a}\right)$.

Vector surface integrals

Vector surface integrals find widespread application in many real-world problems in chemical engineering. As an introductory example, consider the flow of a fluid through a pipe. The velocity of the fluid is given by the vector field, say $\boldsymbol{u}$ and we might want to know what is the total volume of the fluid passing through the pipe per unit time. The volume flow rate is referred to as the flux of the fluid across the surface $\mathcal{S}$ that forms the end of the pipe. Now, if the velocity of the fluid is uniform throughout the pipe (i.e. independent of position), constant with time and moves parallel to the walls of the pipe, then the velocity can be represented by a constant value which implies that the fluid is moving along the pipe as a solid block. If we consider a case where the flow is still parallel to the walls of the pipe but now the velocity depends on position, then $|\boldsymbol{u}|=f(x, y)$ (we are considering a cartesian coordinate system here) and $f$ is some function of $x$ and $y$. Then, the fluid flux, say $Q$, across a small element of a surface $\mathcal{S}$ (giving the change in small amounts $d x$ and $d y$ ) is simply given by $d Q=f(x, y) d S=f(x, y) d x d y$. To determine the total fluid flux therefore, we need to add up all the small contributions which turns out to be defined by the double integral given in Definition 2.8 .

Vector surface integrals are important when we are considering a case in which $\boldsymbol{u}$ and the surface $\mathcal{S}$ are arbitrary; for example, $\mathcal{S}$ may be a rather complicated, curved surface. The ideas are carried forward of course: we still want to consider a small surface, $d S$ and compute the flux of $\boldsymbol{u}$ across $d S$. If $\boldsymbol{u}$ is perpendicular to $d S$ [see Fig. 2.8(a)] then the flux is just $|\boldsymbol{u}| d S$. If, $\boldsymbol{u}$ is not perpendicular to $d S$ [see Fig. 2.8(b)], the total fluid flux through $d S$ is equivalent to the component of $\boldsymbol{u}$ that is perpendicular to $d S$. For this reason, we introduce $\boldsymbol{n}$, which is the unit normal vector to $d S$ [Fig. 2.8(b)].

As previously mentioned, the contribution to the flux through the small surface element is given by the component of $\boldsymbol{u}$ that is perpendicular to $d S$. This is equivalent to the component of $\boldsymbol{u}$ in the direction of the unit normal vector which is $\boldsymbol{u} \cdot \boldsymbol{n}$. Then, the flux across $d S$ is $\boldsymbol{u} \cdot \boldsymbol{n} d S$. The total flux is then given by:

where the integral on the RHS is a standard surface integral.

Generalising Eq. (2.106), we have the following statement: given a vector field, $\boldsymbol{F}$ with a unit normal vector $\boldsymbol{n}$, then the surface integral of $\boldsymbol{F}$ over the surface $\mathcal{S}$ is given by:

(a)

(b)

Figure 2.8: The vector field $\boldsymbol{u}$ across a small surface element $d S$. In (a) the velocity field is perpendicular to $d S$. Subfigure (b) shows the unit normal vecor, $\boldsymbol{n}$ perpendicular to the surface element $d S$.

Finally, we need to discuss the case where the surface is parametrised. Again, we consider the surface $\mathcal{S}$ expressed in terms of two parameters, $u$ and $v$ such that the position vector lying in the surface is given by $\boldsymbol{G}(u, v)$. To evaluate the surface integral, we need an expression for $\boldsymbol{n} \cdot d S$. The surface element $d S$ is given by Eq. (2.104) and the unit normal vector $\boldsymbol{n}$ is given by the cross product of the two vectors, $\boldsymbol{G}_{u}, \boldsymbol{G}_{v}$ (recall that subscript notation here denotes partial differentiation) divided by the magnitude of the cross product of the two vectors, i.e.:

Together with Eq. (2.104), the quantity $\boldsymbol{n} d S$ is:

Computing a vector surface integral

Let $\boldsymbol{G}(u, v)$ be a parametrisation for a surface $\mathcal{S}$ in some region $\mathcal{R}$, where $\mathcal{R}$ is the range of the parameters (i.e. $u$ and $v$ ) that trace the surface $\mathcal{S}$ in the chosen $u v$ plane. Then, the surface integral of the vector field, $\boldsymbol{F}$ is given by:

Next, we look at an example of evaluating vector surface integrals.

Figure 2.9: A point on the cylindrical surface $x^{2}+y^{2}=1$ can be expressed in terms of the parameters $z$ and $\theta$. The figure refers to Example 2.12. Example 2.12 Evaluate the following integral:

where $\mathcal{S}$ is the curved surface of the cylinder $x^{2}+y^{2}=1$ lying between $z=0$ and $z=1$ and $\boldsymbol{u}=(x, z,-y)$.

Solution The two parameters that describe the surface are the height of the cylinder $z$ and the angle $\theta$ around the cylinder as shown in Fig. 2.9. Our parametrisation therefore is given in terms of $\boldsymbol{G}(z, \theta)$ :

where note that we have used the fact that the radius, $r$ on the surface $x^{2}+y^{2}=r^{2}$ is equal to 1 . The partial derivatives of $\boldsymbol{G}$ with respect to the two parameters are:

their cross product is:

Using Eq. (2.106), and applying it to our problem, we obtain a surface integral in $d \theta d z$ where $0 \leq z \leq 1$ (as stated in the problem) and $0 \leq \theta \leq$ $2 \pi$, This yields:

Evaluating the integral gives $\mathcal{I}=\pi$.