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Math 344: Calculus III

15.5 Surface Area

15.5.1 Computing the Area of a Surface ¶

The double integral can be applied to the graph of a function of two variables $z = f(x,y)$ to compute the area of the surface. We will start with computing the area of a surface over a rectangular domain, and then proceed to surfaces over more interesting domains. The mirrors our study of the definition of double integrals and our analysis will be very similar.

We start with a surface $z = f(x,y)$ over a rectangular domain. Then we partition the surface into rectangles. This partitions the surface into sections of surface over each rectangle in the partition.

Surface Over Rectangular Domain Figure 1

The key difference here is that we will approximate a small piece of surface $S_{jk}$ over rectangle $R_{jk}$ using a parallelogram instead of approximating a small volume using a rectangular parallelopiped .

We start by computing the partial derivatives at the lower corner of each rectangle $R_{jk}$ to obtain two linearly independent tangent vectors on the surface $S_{jk}$.

$$ \begin{align*} \require{color} \definecolor{brightblue}{rgb}{.267, .298, .812} \definecolor{darkblue}{rgb}{.08, .18, .28} \definecolor{palepink}{rgb}{1, .73, .8} \definecolor{softmagenta}{rgb}{.99,.34,.86} \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \newcommand{\pypx}[2][x]{\dfrac{\partial #2}{\partial #1}} \newcommand{\dydx}[2][x]{\dfrac{d #2}{d #1}} \newcommand{\deltax}[2][x]{\frac{\Delta #2}{\Delta #1}} \mathbf{r} &= \left\langle 1, 0, \dfrac{\partial f}{\partial x}(x_j,y_k)\right\rangle \\ \\ \mathbf{s} &= \left\langle 0, 1, \dfrac{\partial f}{\partial y}(x_j,y_k)\right\rangle \end{align*} $$
We also scale each tangent vector by the length and width of rectangle $R_{jk}$, $\Delta x_j$ and $\Delta y_k$, respectively. So our tangent vectors become

$$ \begin{align*} \mathbf{r} &= \left\langle \Delta x_j, 0, \dfrac{\partial f}{\partial x}(x_j,y_k)\Delta x_j\right\rangle \\ \\ \mathbf{s} &= \left\langle 0, \Delta y_k, \dfrac{\partial f}{\partial y}(x_j,y_k)\Delta y_k\right\rangle \end{align*} $$

Surface Partial Derivative Vectors Figure 2

15.5.2 Approximation by Parallelograms ¶

We use our vectors $\mathbf{r}$ and $\mathbf{s}$ to construct a parallelogram $T_{jk}$ with sides $\mathbf{r}$ and $\mathbf{s}$.

Surface with Approximating Parallelogram Figure 3

We learned in chapter 12 that the area $\Delta T_{jk}$ of each parallelogram $T_{jk}$ is given by

$$ \text{Area} = \Delta T_{jk} = \left|\mathbf{r}\right|\,\left|\mathbf{s}\right|\sin(\theta) = \left|\mathbf{r}\times\mathbf{s}\right| $$
Now we can compute the cross product of two vectors $\mathbf{r}\times\mathbf{s} = $

$$ \begin{align*} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \Delta x_j & 0 & \dfrac{\partial f}{\partial x}(x_j,y_k)\Delta x_j \\ 0 & \Delta y_k & \dfrac{\partial f}{\partial y}(x_j,y_k)\Delta y_k \end{vmatrix} \\ \\ &= \mathbf{i}\begin{vmatrix} 0 & \dfrac{\partial f}{\partial x}(x_j,y_k)\Delta x_j \\ \Delta y_k & \dfrac{\partial f}{\partial y}(x_j,y_k)\Delta y_k \end{vmatrix} - \mathbf{j}\begin{vmatrix} \Delta x_j & \dfrac{\partial f}{\partial x}(x_j,y_k)\Delta x_j \\ 0 & \dfrac{\partial f}{\partial y}(x_j,y_k)\Delta y_k \end{vmatrix} + \mathbf{k}\begin{vmatrix} \Delta x_j & 0 \\ 0 & \Delta y_k \end{vmatrix} \\ \\ &= \left(-\dfrac{\partial f}{\partial x}(x_j,y_k)\Delta x_j\Delta j_k\right)\mathbf{i} - \left(\dfrac{\partial f}{\partial y}(x_j,y_k)\Delta x_j\Delta y_k\right)\mathbf{j} + \Delta x_j\Delta y_k \\ \\ &= \left[-\dfrac{\partial f}{\partial x}(x_j,y_k)\mathbf{i} - \dfrac{\partial f}{\partial y}(x_j,y_k)\mathbf{j} + \mathbf{k}\right]\Delta x_j\Delta y_k \\ \\ &= \left[-\dfrac{\partial f}{\partial x}(x_j,y_k)\mathbf{i} - \dfrac{\partial f}{\partial y}(x_j,y_k)\mathbf{j} + \mathbf{k}\right]\Delta A_{jk} \end{align*} $$
Now we can compute the magnitude of this vector product

$$ \begin{align*} \Delta T_{jk} &= \left|\mathbf{r}\times\mathbf{s}\right| \\ \\ &= \left(\sqrt{\left(-\dfrac{\partial f}{\partial x}(x_j,y_k)\right)^2 + \left(-\dfrac{\partial f}{\partial y}(x_j,y_k)\right)^2 + 1^2}\right)\Delta A_{jk} \\ \end{align*} $$

15.5.3 Riemann Sum ¶

If we cover each surface $S_{jk}$ with parallelogram $T_{jk}$

Surface Covered with Approximating Parallelograms Figure 4

and add up the areas $\Delta T_{jk}$ of each parallelogram we obtain a Riemann sum.

$$ \displaystyle\sum_{j=1}^m\displaystyle\sum_{k=1}^n \left(\sqrt{f_x(x_j,y_k)^2 + f_y(x_j,y_k)^2 + 1}\right)\Delta A_{jk} $$
We can see in these plots that the approximating parallelograms $T_{jk}$ very close to each surface $S_{jk}$ for a smooth surface.

Surface with Approximating Parallelograms Figure 5

Surface with Approximating Parallelograms Figure 6

15.5.4 Double Integral Formula ¶

If we compute the limit as $m\rightarrow\infty$ and $n\rightarrow\infty$, we obtain a double integral that evaluates to the surface area.

$$ \begin{align*} A(S) &= \displaystyle\iint_D \sqrt{\left[f_x(x,y)\right]^2 + \left[f_y(x,y)\right]^2 + 1}\,dA \\ \\ &= \displaystyle\iint_D \sqrt{\left(\dfrac{\partial z}{\partial x}\right)^2 + \left(\dfrac{\partial z}{\partial y}\right)^2 + 1}\,dA \end{align*} $$
Notice how similar this integral formula for surface area is to the integral formula for arc length.

$$ L = \displaystyle\int_a^b \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2}\,dx $$

15.1.5 Examples ¶

Example 1 ¶

Let region $D$ be the triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$. Compute the area of the surface given by $z = 4 - x^2 - 2y$.

Computing an iterated integral one obtains

$$ \begin{align*} A(S) &= \displaystyle\iint_D \sqrt{\left[f_x(x,y)\right]^2 + \left[f_y(x,y)\right]^2 + 1}\,dA \\ \\ &= \displaystyle\int_0^1\displaystyle\int_0^x \sqrt{(-2x)^2 + (-2)^2 + 1}\,dy\,dx \\ \\ &= \displaystyle\int_0^1\displaystyle\int_0^x \sqrt{4x^2 + 5}\,dy\,dx \\ \\ &= \displaystyle\int_0^1 x\sqrt{4x^2 + 5}\,dx\qquad\left|\begin{array}{l} u = 4x^2 + 5 \\ du = 8x\,dx \end{array}\right. \\ \\ &= \dfrac{1}{8} \displaystyle\int_5^9 \sqrt{u}\,du = \dfrac{1}{8}\,\left[\dfrac{2}{3}\,u^{3/2}\right]_5^9 \\ \\ &= \dfrac{1}{12}\,\left[ 9^{3/2} - 5^{3/2} \right] = \dfrac{1}{12}\left[27 - 5\sqrt{5}\right] \end{align*} $$

Example 2 ¶

Find the area of the surface $z = 4 - x^2 - y^2$ that lies above the disk of radius 2.

Computing an iterated integral one obtains

$$ \begin{align*} A(S) &= \displaystyle\iint_D \sqrt{\left[f_x(x,y)\right]^2 + \left[f_y(x,y)\right]^2 + 1}\,dA \\ \\ &= \displaystyle\iint_D \sqrt{(-2x)^2 + (-2y)^2 + 1}\,dA \\ \\ &= \displaystyle\int_0^{2\pi}\,\displaystyle\int_0^2 \sqrt{4r^2 + 1}\,r\,dr\,d\theta \qquad\left|\begin{array}{l} u = 4r^2 + 1 \\ du = 8r\,dr \end{array}\right. \\ \\ &= \dfrac{1}{8}\,\displaystyle\int_0^{2\pi}\,d\theta\cdot\displaystyle\int_1^{17}\sqrt{u}\,du = \dfrac{1}{8}\cdot\left[\,\theta\,\right]_0^{2\pi}\cdot\left[\,\dfrac{2}{3}u^{3/2}\,\right]_1^{17} \\ \\ &= \dfrac{1}{8}\cdot 2\pi\cdot\dfrac{2}{3}\,\left[ 17^{3/2} - 1^{3/2}\right] = \dfrac{\pi}{6}\,\left(17\sqrt{17} - 1\right) \end{align*} $$

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