In Section 7.4 we used definite integrals to compute the arc length of plane curves of the form $y=f(x)\text{.}$ We later extended these ideas to compute the arc length of plane curves defined by parametric or polar equations.

The natural extension of the concept of “arc length over an interval” to surfaces is “surface area over a region.”

Consider the surface $z=f(x,y)$ over a region $R$ in the $x$-$y$ plane, shown in Figure 13.5.1(a). Because of the domed shape of the surface, the surface area will be greater than that of the area of the region $R\text{.}$ We can find this area using the same basic technique we have used over and over: we'll make an approximation, then using limits, we'll refine the approximation to the exact value.

As done to find the volume under a surface or the mass of a lamina, we subdivide $R$ into $n$ subregions. Here we subdivide $R$ into rectangles, as shown in the figure. One such subregion is outlined in the figure, where the rectangle has dimensions $\dx_i$ and $\dy_i\text{,}$ along with its corresponding region on the surface.

In part (b) of the figure, we zoom in on this portion of the surface. When $\dx_i$ and $\dy_i$ are small, the function is approximated well by the tangent plane at any point $(x_i,y_i)$ in this subregion, which is graphed in part (b). In fact, the tangent plane approximates the function so well that in this figure, it is virtually indistinguishable from the surface itself! Therefore we can approximate the surface area $S_i$ of this region of the surface with the area $T_i$ of the corresponding portion of the tangent plane.

This portion of the tangent plane is a parallelogram, defined by sides $\vec u$ and $\vec v\text{,}$ as shown. One of the applications of the cross product from Section 10.4 is that the area of this parallelogram is $\norm{\vec u\times \vec v}\text{.}$ Once we can determine $\vec u$ and $\vec v\text{,}$ we can determine the area.

$\vec u$ is tangent to the surface in the direction of $x\text{,}$ therefore, from Section 12.7, $\vec u$ is parallel to $\la 1,0,f_x(x_i,y_i)\ra\text{.}$ The $x$-displacement of $\vec u$ is $\dx_i\text{,}$ so we know that $\vec u = \dx_i\la 1,0,f_x(x_i,y_i)\ra\text{.}$ Similar logic shows that $\vec v = \dy_i\la 0,1,f_y(x_i,y_i)\ra\text{.}$ Thus: \begin{align*} \text{ surface area $S_i$ } \amp \approx \text{ area of $T_i$ }\\ \amp = \norm{\vec u\times \vec v}\\ \amp = \norm{\dx_i\la 1,0,f_x(x_i,y_i)\ra\times\dy_i\la 0,1,f_y(x_i,y_i)\ra}\\ \amp =\sqrt{1+f_x(x_i,y_i)^2+f_y(x_i,y_i)^2}\dx_i\dy_i. \end{align*}

Note that $\dx_i\dy_i = \Delta A_i\text{,}$ the area of the $i^{\,\text{ th } }$ subregion.

Summing up all $n$ of the approximations to the surface area gives \begin{equation*} \text{ surface area over $R$ } \approx \sum_{i=1}^n \sqrt{1+f_x(x_i,y_i)^2+f_y(x_i,y_i)^2}\Delta A_i. \end{equation*}

Once again take a limit as all of the $\dx_i$ and $\dy_i$ shrink to 0; this leads to a double integral.

##### Definition13.5.4Surface Area

Let $z=f(x,y)$ where $f_x$ and $f_y$ are continuous over a closed, bounded region $R\text{.}$ The surface area $S$ over $R$ is \begin{align*} S \amp = \iint_R \ dS\\ \amp =\iint_R \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\ dA. \end{align*}

as done before, we think of “$\iint_R\ dS$” as meaning “sum up lots of little surface areas over $R\text{.}$”

The concept of surface area is defined here, for while we already have a notion of the area of a region in the plane, we did not yet have a solid grasp of what “the area of a surface in space” means.

We test this definition by using it to compute surface areas of known surfaces. We start with a triangle.

##### Example13.5.5Finding the surface area of a plane over a triangle

Let $f(x,y) = 4-x-2y\text{,}$ and let $R$ be the region in the plane bounded by $x=0\text{,}$ $y=0$ and $y=2-x/2\text{,}$ as shown in Figure 13.5.6. Find the surface area of $f$ over $R\text{.}$

Solution

It is “common knowledge” that the surface area of a sphere of radius $r$ is $4\pi r^2\text{.}$ We confirm this in the following example, which involves using our formula with polar coordinates.

##### Example13.5.7The surface area of a sphere.

Find the surface area of the sphere with radius $a$ centered at the origin, whose top hemisphere has equation $f(x,y)=\sqrt{a^2-x^2-y^2}\text{.}$

Solution

The inner integral in Equation <<Unresolved xref, reference "eq_exsurfacearea2"; check spelling or use "provisional" attribute>> is an improper integral, as the integrand of $\ds \int_0^ar\sqrt{\frac{a^2}{a^2-r^2}}\ dr$ is not defined at $r=a\text{.}$ To properly evaluate this integral, one must use the techniques of Section 6.8.

The reason this need arises is that the function $f(x,y) = \sqrt{a^2-x^2-y^2}$ fails the requirements of Definition 13.5.4, as $f_x$ and $f_y$ are not continuous on the boundary of the circle $x^2+y^2=a^2\text{.}$

The computation of the surface area is still valid. The definition makes stronger requirements than necessary in part to avoid the use of improper integration, as when $f_x$ and/or $f_y$ are not continuous, the resulting improper integral may not converge. Since the improper integral does converge in this example, the surface area is accurately computed.

##### Example13.5.8Finding the surface area of a cone

The general formula for a right cone with height $h$ and base radius $a$ is \begin{equation*} \ds f(x,y) = h-\frac{h}a\sqrt{x^2+y^2}, \end{equation*} shown in Figure 13.5.9. Find the surface area of this cone.

Solution

Note that once again $f_x$ and $f_y$ are not continuous on the domain of $f\text{,}$ as both are undefined at $(0,0)\text{.}$ (A similar problem occurred in the previous example.) Once again the resulting improper integral converges and the computation of the surface area is valid.

##### Example13.5.10Finding surface area over a region

Find the area of the surface $f(x,y) = x^2-3y+3$ over the region $R$ bounded by $-x\leq y\leq x\text{,}$ $0\leq x\leq 4\text{,}$ as pictured in Figure 13.5.11.

Solution

In practice, technology helps greatly in the evaluation of such integrals. High powered computer algebra systems can compute integrals that are difficult, or at least time consuming, by hand, and can at the least produce very accurate approximations with numerical methods. In general, just knowing how to set up the proper integrals brings one very close to being able to compute the needed value. Most of the work is actually done in just describing the region $R$ in terms of polar or rectangular coordinates. Once this is done, technology can usually provide a good answer.

We have learned how to integrate integrals; that is, we have learned to evaluate double integrals. In the next section, we learn how to integrate double integrals — that is, we learn to evaluate triple integrals, along with learning some uses for this operation.

# Subsection13.5.1Exercises

Terms and Concepts

In the following exercises, set up the iterated integral that computes the surface area of the given surface over the region $R\text{.}$