Computing Surface Areas with Double Integrals

Let $z = f(x, y)$ be a two variable real-valued function that represents the surface $S$, and suppose that $f$ has continuous first partial derivatives and assume that $f(x, y) ≥ 0$ and suppose that the domain $D$ of the surface $S$ is a rectangle.

Now let's take the domain $D$ and divide it into sub-rectangles of equal width and equal length and with area $\Delta A = \Delta x \Delta y$. Let the point $(x_i, y_j)$ be the corner point in the sub-rectangle $R_{ij}$ that is closest to the origin. Then the point $P_{ij} (x_i, y_j, f(x_i, y_j))$ lies on the surface $S$. Recall that the tangent plane at a point best approximates the surface near that point.

Let $T_{ij}$ be the parallelogram portion of the tangent plane at $P_{ij}$ that lies directly above the sub-rectangle $R_{ij}$. The area of $T_{ij}$ is $\Delta T_{ij}$ which approximates the surface area directly above $R_{ij}$. Therefore, we have that if $A(S)$ represents the surface area of $S$ then:

(1)

\begin{align} \quad A(S) \approx \sum_{i=1}^{m} \sum_{j=1}^{n} \Delta T_{ij} \end{align}

As the number of rectangles increases, our approximation of the surface area gets better and better and hence:

(2)

\begin{align} \quad A(S) = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} \Delta T_{ij} \end{align}

Now let $\vec{a}$ and $\vec{b}$ be vectors that start at the point $P_{ij}$ and that run along the sides of the parallelogram that determine the area approximation $T_{ij}$.

Note that $\frac{\partial f}{\partial x} (x_i, y_j)$ is the slope of the tangent line through $P_{ij}$ and in the direction of $\vec{a}$ while $\frac{\partial f}{\partial y} (x_i, y_j)$ is the slope of the tangent line through $P_{ij}$ and in the direction of $\vec{b}$, and so we can write these vectors as:

(3)

\begin{align} \quad \vec{a} = \Delta x \vec{i} + 0 \vec{j} + \frac{\partial f}{\partial x} (x_i, y_j) \Delta x \vec{k} \quad , \quad \vec{b} = 0 \vec{i} + \Delta y \vec{j} + \frac{\partial f}{\partial y} (x_i, y_j) \Delta y \vec{k} \end{align}

Recall that the area of this parallelogram approximation is $\mid \vec{a} \times \vec{b} \mid$. We first compute $\vec{a} \times \vec{b}$ as follows:

(4)

\begin{align} \quad \quad \vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \Delta x & 0 & \frac{\partial f}{\partial x} (x_i, y_j) \Delta x\\ 0 & \Delta y & \frac{\partial f}{\partial y} (x_i, y_j) \Delta y \end{vmatrix} = - \frac{\partial f}{\partial x} (x_i, y_j) \Delta x \Delta y \vec{i} - \frac{\partial f}{\partial y} (x_i, y_j) \Delta x \Delta y \vec{j} + \Delta x \Delta y \vec{k} = \left [ - \frac{\partial f}{\partial x}(x_i, y_j) \vec{i} - \frac{\partial f}{\partial y}(x_i, y_j) \vec{j} + \vec{k} \right ] \Delta A \end{align}

Therefore the area of the parallelogram is $T_{ij} = \mid \vec{a} \times \vec{b} \mid$:

(5)

\begin{align} \quad \Delta T_{ij} = \mid \vec{a} \times \vec{b} \mid = \sqrt{ \left [ \frac{\partial f}{\partial x}(x_i, y_j) \right ]^2 + \left [ \frac{\partial f}{\partial y}(x_i, y_j) \right ]^2 + 1} \Delta A \end{align}

Therefore the area $A(S)$ can be given by:

(6)

\begin{align} A(S) = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} \Delta T_{ij} = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} \sqrt{ \left [ \frac{\partial f}{\partial x}(x_i, y_j) \right ]^2 + \left [ \frac{\partial f}{\partial y}(x_i, y_j) \right ]^2 + 1} \Delta A \end{align}

Of course, we can rewrite the double Riemann sum above as a double integral:

(7)

\begin{align} \quad A(S) = \iint_D \sqrt{ \left [ \frac{\partial f}{\partial x}\right ]^2 + \left [ \frac{\partial f}{\partial y} \right ]^2 + 1} \: dA \end{align}