Surface Integrals

Recall that if $y = f(x)$ is a continuous single variable function for $a ≤ x ≤ b$ then $\int_a^b f(x) \: dx$ is the integral of $f$ over the line segment whose initial point is $(a, 0)$ and whose terminal point is $(b, 0)$. We extended this concept further to integrating a function of two variables over a curve and we called this type of integral, a line integral.

When dealing with double integrals, we considered a region of $\mathbb{R}^2$ and integrated over it. Suppose that now we want to integrate a three variable bounded function $w = f(x, y, z)$ over a smooth finite smooth surface $\delta$ in $\mathbb{R}^3$. To do so, let's first take the surface $\delta$ in $\mathbb{R}^3$ and subdivide into $n$ pieces $\delta_i$ that are approximately planar and that do not overlap each other, and suppose that each piece has area $\Delta s_i$. Choose arbitrary points $(x_i^*, y_i^*, z_i^*)$ in each $\delta_i$. We then form a Riemann sum $R_n$ which approximates the value of the integral of $f$ over this surface:

(1)

\begin{align} \quad R_n = \sum_{i=1}^n f(x_i^*, y_i^*, z_i^*) \Delta s_i \end{align}

Now if the limit of $R_n$ as the diameters of each $\delta_i$ approach zero as the number of non overlapping subdivisions of $\delta$, $n$, approaches infinity, then this is what we defined as the surface integral of $f$ over $\delta$.

(2)

\begin{align} \quad \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*, y_i^*, z_i^*) \Delta s_i = \iint_{\delta} f(x, y, z) \: dS \end{align}

Definition: Let $\delta$ be a smooth surface with finite area in $\mathbb{R}^3$ and let $w = f(x, y, z)$ be a three variable real-valued function that is bounded on $\delta$. Then the Surface Integral of $f$ over $\delta$ is $\iint_{\delta} f(x, y, z) \: dS$.

Let's try to come up with a formula to compute these surface integrals. Consider a smooth parametric surface that generates the surface $\delta$ and is given by $\vec{r}(u, v)$ whose domain is $R$. Let $(u_0, v_0)$ be a point that is containing in the interior of $R$. Then $\vec{r}(u_0, v)$ and $\vec{r}(u, v_0)$ are both curves on the surface $\delta$. Also consider the curves $\vec{r}(u_0 + du, v)$ and $\vec{r}(u, v_0 + dv)$ where $du$ and $dv$ represent the change in $u$ and $v$ respectively.

Now let's look at the tangent vectors of the curves $\vec{r}(u_0, v)$ and $\vec{r}(u, v_0)$ at their point of intersection $\vec{r_0} = (u_0, v_0)$. The tangent vector at this point of intersection on the curve $\vec{r}(u_0, v)$ with length $dv$ is $\frac{\partial \vec{r}}{\partial v} dv$. Similarly, the tangent vector at this point of intersection on the curve $\vec{r}(u, v_0)$ with length $du$ is $\frac{\partial \vec{r}}{\partial u} du$.

Provided that these two tangent vectors are not parallel, then we can take their cross product to obtain a vector that is normal to the surface $\delta$ at $\vec{r_0} (u_0, v_0)$. Note that $\frac{\partial \vec{r}}{\partial u} = \left ( \frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u} \right ) = \frac{\partial x}{\partial u} \vec{i} + \frac{\partial y}{\partial u} \vec{j} + \frac{\partial z}{\partial u} \vec{k}$ and $\frac{\partial \vec{r}}{\partial v} = \left ( \frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v} \right ) = \frac{\partial x}{\partial v} \vec{i} + \frac{\partial y}{\partial v} \vec{j} + \frac{\partial z}{\partial v} \vec{k}$. Thus the cross product between these two vectors is:

(3)

\begin{align} \quad \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u}\\ \frac{\partial x}{\partial v} & \frac{\partial x}{\partial v} & \frac{\partial z}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \end{vmatrix} \vec{i} - \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial z}{\partial v} \end{vmatrix} \vec{j} + \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{vmatrix} \vec{k} = \frac{\partial (y, z)}{\partial (u, v)} \vec{i} + \frac{\partial (z, x)}{\partial (u,v)} \vec{j} + \frac{\partial (x, y)}{\partial (u, v)} \vec{k} \end{align}

The area element $dS$ can be obtained as the area spanned by these tangent vectors $\frac{\partial \vec{r}}{\partial u} du$ and $\frac{\partial \vec{r}}{\partial v} dv$. Recall that the area spanned by two vectors (forming a parallelogram) in $\mathbb{R}^3$ is equal to the norm of the cross product of these two vectors, that is:

(4)

\begin{align} \quad dS = \biggr \| \frac{\partial \vec{r}}{\partial u} du \times \frac{\partial \vec{r}}{\partial v} dv \biggr \| = \biggr \| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \biggr \| \: du \: dv = \sqrt{ \left ( \frac{\partial (y, z)}{\partial (u, v)} \right )^2 + \left ( \frac{\partial (z, x)}{\partial (u, v)} \right )^2 + \left ( \frac{\partial (x, y)}{\partial (u, v)} \right )^2 } \: du \: dv \end{align}

Therefore the surface integral of the function $f$ over the surface $\delta$ defined parametrically by $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$ for $(u, v) \in D$ is given by the following formula:

(5)

\begin{align} \quad \quad \iint_{\delta} f(x, y, z) \: dS = \iint_D f(\vec{r}(u, v)) \biggr \| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} \biggr \| \: du \: dv = \iint_D f(x(u, v), y(u, v), z(u, v)) \sqrt{ \left ( \frac{\partial (y, z)}{\partial (u, v)} \right )^2 + \left ( \frac{\partial (z, x)}{\partial (u, v)} \right )^2 + \left ( \frac{\partial (x, y)}{\partial (u, v)} \right )^2 } \: du \: dv \end{align}

Geometrically Obtaining the Surface Area Element dS

From above, we see that the surface area element may sometimes be tedious to obtain. Sometimes we will be able to geometrically obtain the surface area element. Suppose that the surface $\delta$ has a one-to-one projection onto the $xy$-plane, that is, a one-to-one projection onto the plane $z = 0$. We will then have that the area element $dS$ can be expressed as:

(6)

\begin{align} \quad dS = \frac{1}{\mid \cos \gamma \mid} \: dx \: dy = \frac{\| n \|}{\| n \cdot k \|} \: dx \: dy \end{align}

The angle $\gamma$ is the angle between a vector $\vec{n}$ that is normal to the surface $\delta$ and $\vec{k} = (0, 0, 1)$. Now note that if $z = g(x, y)$ represents our surface $\delta$ then we can rewrite this equation as $G(x, y, z) = z - g(x, y) = 0$, and furthermore, if $\nabla G(x, y, z)$ is nonzero, then $\nabla G(x, y, z)$ will be normal to $\delta$ (provided that $G$ has continuous first partial derivatives), so we can let:

(7)

\begin{align} \quad \vec{n} = \nabla G(x, y, z) \end{align}

We also note that:

(8)

\begin{align} \quad \vec{n} \cdot \vec{k} = \nabla G(x, y, z) \cdot \vec{k} = \left ( \frac{\partial G}{\partial x}, \frac{\partial G}{\partial y}, \frac{\partial G}{\partial z} \right ) \cdot (0, 0, 1) = \frac{\partial G}{\partial z} \end{align}

Therefore the formula for our surface integral will be:

(9)

\begin{align} \quad \iint_{\delta} f(x, y, z) \: dS = \iint_D f(x, y, g(x, y)) \frac{\| \nabla G(x, y, z) \|}{\biggr \| \frac{\partial G}{\partial z} \biggr \|} \: dx \: dy \end{align}