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.
Consider the surface \(z=f(x,y)\) over a region \(R\) in the \(xy\)-plane, shown in Figure 14.5.1.(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.
A domain is shown in the \(xy\) plane; it appears to be bounded by a circle of radius \(1\text{,}\) centered at \((1,0,0)\text{.}\) The domain is divided into many small rectangles, illustrating a partition like the ones used to define the double integral.
The grid of rectangles in the domain corresponds to a grid of curves on the surface. One of the rectangles in the domain is highlighted, along with the corresponding region on the surface.
This image is a zoomed-in view of the rectangular region that was highlighted on the surface in Figure 14.5.1.(a). Below the surface, we see a single rectangle from the partition of the domain. The sides of this rectangle are marked with lengths \(\Delta x_i\) and \(\Delta y_j\text{.}\)
The rectangle in the \(xy\) plane corresponds to a small, rectangular patch on the surface. This patch is approximated by a piece of the tangent plane at one point, spanned by vectors \(\vec{u}\) and \(\vec{v}\text{.}\) These vectors are labeled, but the point is not.
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 \(\dy_j\text{,}\) along with its corresponding region on the surface.
In part Figure 14.5.1.(b) of the figure, we zoom in on this portion of the surface. When \(\dx_i\) and \(\dy_j\) are small, the function is approximated well by the tangent plane at any point \((x_i,y_j)\) in this subregion, which is graphed in part Figure 14.5.1.(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 \(S_{ij}\) of this region of the surface with the area \(T_{ij}\) 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 11.4 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.
The vector \(\vec u\) is tangent to the surface in the direction of \(x\text{,}\) therefore, from Section 13.7, \(\vec u\) is parallel to \(\la 1,0,f_x(x_i,y_j)\ra\text{.}\) The \(x\)-displacement of \(\vec u\) is \(\dx_i\text{,}\) so we know that \(\vec u = \dx_i\la 1,0,f_x(x_i,y_j)\ra\text{.}\) Similar logic shows that \(\vec v = \dy_j\la 0,1,f_y(x_i,y_j)\ra\text{.}\) Thus:
Summing up all \(n\) of the approximations to the surface area gives
\begin{equation*}
\text{surface area over \(R\)} \approx \sum_{i=1}^m\sum_{j=1}^n \sqrt{1+f_x(x_i,y_j)^2+f_y(x_i,y_j)^2}\Delta A_{ij}\text{.}
\end{equation*}
Example14.5.3.Finding 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 14.5.4. Find the surface area of \(f\) over \(R\text{.}\)
Three-dimensional coordinate axes are drawn in space. In the \(xy\) plane there is a triangle that illustrates the domain for this problem. The triangle is bounded by the \(x\) and \(y\) axes, and the line from \((4,0,0)\) to \((0,2,0)\text{.}\)
This line also happens to be where the plane \(z=4-x-2y\) intersects the \(xy\) plane, so it makes up one of the three sides of the triangle on the surface for this problem. The other two sides are the intersections of \(z=4-x-2y\) with the \(xz\) and \(yz\) planes.
We follow Definition 14.5.2. We start by noting that \(f_x(x,y) = -1\) and \(f_y(x,y) = -2\text{.}\) To define \(R\text{,}\) we use bounds \(0\leq y\leq 2-x/2\) and \(0\leq x\leq 4\text{.}\) Therefore
Because the surface is a triangle, we can figure out the area using geometry. Considering the base of the triangle to be the side in the \(xy\)-plane, we find the length of the base to be \(\sqrt{20}\text{.}\) We can find the height using our knowledge of vectors: let \(\vec u\) be the side in the \(xz\)-plane and let \(\vec v\) be the side in the \(xy\)-plane. The height is then \(\norm {\vec u - \proj{u}{v}} = 4\sqrt{6/5}\text{;}\) this is indicated by the additional line drawn in Figure 14.5.4. Geometry states that the area is thus
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.
The region \(R\) that we are integrating over is bounded by the circle, centered at the origin, with radius \(a\text{:}\)\(x^2+y^2=a^2\text{.}\) Because of this region, we are likely to have greater success with our integration by converting to polar coordinates. Using the substitutions \(x=r\cos(\theta)\text{,}\)\(y=r\sin(\theta)\text{,}\)\(dA = r\, dr\, d\theta\) and bounds \(0\leq\theta\leq2\pi\) and \(0\leq r\leq a\text{,}\) we have:
A circular cone, with vertex on the \(z\) axis at a point \((0,0,h)\text{.}\) The cone opens downward toward the \(xy\) plane, and ends at the \(xy\) plane with a circular base of radius \(a\text{.}\)
Since we are integrating over the disk bounded by \(x^2+y^2=a^2\text{,}\) we again use polar coordinates. Using the standard substitutions, our integrand becomes
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 14.5.9.
The surface is plotted against the usual three-dimensional coordinate axes. In the \(xy\) plane we see the triangular outline of the domain for this problem. The vertices of this triangle are at \((4,4,0)\text{,}\)\((4,-4,0)\text{,}\) and \((0,0,0)\text{.}\)
On the surface, the curves corresponding to the three sides of this triangle are shown, illustrating the triangular region whose surface area is computed in this example.
As with integrals describing arc length, double integrals describing surface area are in general hard to evaluate directly because of the square-root. This particular integral can be easily evaluated, though, with judicious choice of our order of integration.
Integrating with order \(dx\, dy\) requires us to evaluate \(\int \sqrt{10+4x^2}\, dx\text{.}\) This can be done, though it involves Integration By Parts and \(\sinh^{-1}(x)\text{.}\) Integrating with order \(dy\, dx\) has as its first integral \(\int \sqrt{10+4x^2}\, dy\text{,}\) which is easy to evaluate: it is simply \(y\sqrt{10+4x^2}+C\text{.}\) So we proceed with the order \(dy\, dx\text{;}\) the bounds are already given in the statement of the problem.
\begin{align*}
\iint_R\sqrt{10+4x^2}\, dA \amp = \int_0^4\int_{-x}^x\sqrt{10+4x^2}\, dy \, dx\\
\amp = \int_0^4\left.\big(y\sqrt{10+4x^2}\big)\right|_{-x}^x dx\\
\amp =\int_0^4\big(2x\sqrt{10+4x^2}\big)\, dx.\\
\end{align*}
So while the region \(R\) over which we integrate has an area of 16 square units, the surface has a much greater area as its \(z\)-values change dramatically over \(R\text{.}\)
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.
Let \(f(x,y)\) be a function defined over a region \(R\) and let \(g(x,y)=2f(x,y)\text{.}\) Why is the surface area of \(z=g(x,y)\) over \(R\) not twice the surface area of \(z=f(x,y)\) over \(R\text{?}\)
In the following exercises, set up the iterated integral that computes the surface area of the graph of the given function over the region \(R\text{.}\)
The surface \(z=\sin(x)\cos(y)\) is plotted over a rectangular domain. The domain is large enough that several of the peaks and valleys typical of this surface can be seen.
Note that the image is somewhat decorative in this case: the formula for surface area can be applied to the given function without knowing the appearance of the surface.
The surface \(z=\frac{1}{x^2+y^2+1}\) has rotational symmetry about the \(z\) axis. It resembles a steep mountain with a smooth peak, or perhaps the hat of a witch.
A standard hyperbolic paraboloid, or saddle surface. The saddle point is at the origin. The surface curves upward along the \(x\) axis, and downward along the \(y\) axis.
A graph of the function \(f(x,y) = \frac{1}{e^{x^2}+1}\) in three dimensions. The surface is mostly flat, with a ridge along the \(y\) axis of height \(1/2\text{.}\)
\(z = 10-2\sqrt{x^2+y^2}\) over the region \(R\) bounded by the circle \(x^2+y^2=25\text{.}\) (This is the cone with height 10 and base radius 5; be sure to compare your result with the known formula.)
Find the surface area of the sphere with radius 5 by doubling the surface area of \(f(x,y) = \sqrt{25-x^2-y^2}\) over the region \(R\) bounded by the circle \(x^2+y^2=25\text{.}\) (Be sure to compare your result with the known formula.)
Find the surface area of the ellipse formed by restricting the plane \(f(x,y) = cx+dy+h\) to the region \(R\) bounded by the circle \(x^2+y^2=1\text{,}\) where \(c\text{,}\)\(d\) and \(h\) are some constants. Your answer should be given in terms of \(c\) and \(d\text{;}\) why does the value of \(h\) not matter?
