From a Graph to a Surface Area Integral
A surface given by a graph \(z=f(x,y)\) can be viewed as a deformed copy of its domain in the \(xy\)-plane. Where the graph tilts, a small piece of the surface has more area than its projection onto the plane. The surface-area factor measures exactly that local increase.
For a continuously differentiable function, the factor is determined by the two partial derivatives. The practical task is then to integrate it over the region \(D\) in the plane that supports the graph. Rectangles, disks, and annuli are especially useful examples because their bounds are direct or have a simple polar form.
The expression under the square root is at least \(1\), so the surface area is at least the area of its projection. Equality holds when both partial derivatives vanish throughout the region, as for a horizontal graph.
Why the Surface-Area Factor Has This Form
A short derivation explains the formula. Parametrize the graph by the coordinates of its projection: \(\mathbf{r}(x,y)=(x,y,f(x,y))\). The two tangent vectors in the coordinate directions are \(\mathbf{r}_x=(1,0,f_x)\) and \(\mathbf{r}_y=(0,1,f_y)\). The area of the small parallelogram they span is the magnitude of their cross product.
Thus a small projected area \(dA\) corresponds locally to surface area \(\sqrt{1+f_x^2+f_y^2}\,dA\). Adding these local contributions over \(D\), by integration, gives the stated formula. The differentiability condition ensures that the tangent vectors and local area factor are defined throughout the region.
Worked Example: A Curved Graph Over a Rectangle
Worked Example: A Curved Graph Over a Rectangle
Find the area of the graph \(z=x^2\) above the rectangle \(D=[0,1]\times[0,2]\).
First calculate the partial derivatives: \(f_x=2x\) and \(f_y=0\). The area factor is therefore \(\sqrt{1+4x^2}\), independent of \(y\). Using the rectangular bounds gives
An antiderivative is \(F(x)=\frac{x}{2}\sqrt{1+4x^2}+\frac14\operatorname{arsinh}(2x)\). Indeed, differentiating its first term gives \(\frac12\sqrt{1+4x^2}+\frac{2x^2}{\sqrt{1+4x^2}}\), and differentiating its second gives \(\frac{1}{2\sqrt{1+4x^2}}\). Their sum is \(\frac{1+4x^2}{\sqrt{1+4x^2}}=\sqrt{1+4x^2}\), as required. Since \(\operatorname{arsinh}(2)=\ln(2+\sqrt5)\), evaluation yields
This is greater than the projected rectangle area \(2\), as expected. The derivative check verifies the antiderivative used in the calculation.
Disks and Polar Coordinates
For a disk centered at the origin, polar coordinates describe the region as \(0\le r\le R\), \(0\le\theta\le2\pi\). The area element is \(dA=r\,dr\,d\theta\), where the factor \(r\) is essential. If the graph-area factor depends only on \(r\), the angular integration contributes a factor of \(2\pi\).
Worked Example: A Paraboloid Over a Disk
Find the area of \(z=x^2+y^2\) above the disk \(x^2+y^2\le1\). The derivatives are \(f_x=2x\) and \(f_y=2y\), so the area factor is \(\sqrt{1+4x^2+4y^2}\). In polar coordinates \(x^2+y^2=r^2\), and the disk has \(0\le r\le1\). Therefore
Substitute \(u=1+4r^2\), so \(du=8r\,dr\). The radial integral is \(\frac18\int_1^5u^{1/2}\,du=\frac1{12}[u^{3/2}]_1^5=\frac{5\sqrt5-1}{12}\). Equivalently, an antiderivative of \(r\sqrt{1+4r^2}\) is \((1+4r^2)^{3/2}/12\): its derivative is \((1/12)(3/2)(8r)\sqrt{1+4r^2}=r\sqrt{1+4r^2}\). Thus
The projected disk has area \(\pi\), and the result exceeds \(\pi\), consistent with the surface being curved upward.
Symmetry Can Reduce a Double Integral
Symmetry helps when it makes the integrand simpler, or makes the same integral repeat over equal angular or spatial portions of the domain. It does not mean that an arbitrary surface-area integral can be replaced by an area times one convenient value: the integrand must actually be constant for that shortcut. A particularly useful situation occurs when \(f_x^2+f_y^2\) depends only on \(r\), even if the function \(f\) itself is not radial.
Worked Example: Symmetry for a Saddle Graph Over a Disk
Find the area of \(z=xy\) above the unit disk. Its partial derivatives are \(f_x=y\) and \(f_y=x\). Hence the area factor is \(\sqrt{1+x^2+y^2}=\sqrt{1+r^2}\). Although the graph \(z=xy\) changes sign in different quadrants, its area factor depends only on the distance \(r\) from the origin. The disk and integrand are rotationally symmetric, so the angular integral is \(2\pi\):
Set \(u=1+r^2\), so \(du=2r\,dr\). Then \(\int_0^1r\sqrt{1+r^2}\,dr=\frac12\int_1^2u^{1/2}\,du=\frac13(2^{3/2}-1)=\frac13(2\sqrt2-1)\). Consequently,
To check the radial integration, differentiate \((1+r^2)^{3/2}/3\); its derivative is \(r\sqrt{1+r^2}\), the integrand. The result is larger than the unit disk's projected area \(\pi\).
Annuli: Subtract the Inner Radius, Not the Integrand
An annulus centered at the origin, with inner radius \(a\) and outer radius \(b\), is described by \(a\le r\le b\) and \(0\le\theta\le2\pi\), where \(0\le a<b\). For a radial area factor \(g(r)\), its graph area is \(2\pi\int_a^b g(r)r\,dr\). The radial limits exclude the hole; they do not call for subtracting two integrands.
Worked Example: A Paraboloid Over an Annulus
Find the area of \(z=x^2+y^2\) above the annulus \(1\le x^2+y^2\le4\). As before, the area factor is \(\sqrt{1+4r^2}\). Now the radial limits are \(1\le r\le2\), so
Using the antiderivative \((1+4r^2)^{3/2}/12\), whose derivative was checked above, evaluate at both endpoints:
The projected annulus has area \(\pi(2^2-1^2)=3\pi\), while the computed surface area is larger. The bounds correctly omit the central disk.
A Reliable Setup and a Common Pitfall
Compute \(f_x\) and \(f_y\), then form \(\sqrt{1+f_x^2+f_y^2}\).
Use ordinary coordinate bounds for a rectangle, or polar bounds for a disk or centered annulus when they simplify the integrand.
In polar coordinates, use \(dA=r\,dr\,d\theta\), not \(dr\,d\theta\).
Verify the antiderivative by differentiation and compare the result with the projected area.
A common mistake is to forget the polar factor \(r\). For the unit-disk paraboloid example, omitting it would produce \(2\pi\int_0^1\sqrt{1+4r^2}\,dr\), which is not the surface-area integral: a polar coordinate strip of radial width \(dr\) has area proportional to \(r\,dr\). The correct setup is \(2\pi\int_0^1r\sqrt{1+4r^2}\,dr\). Always transform the area element along with the integrand and bounds.
Another useful check is that \(\sqrt{1+f_x^2+f_y^2}\ge1\). Therefore the graph area cannot be smaller than the area of its projection. This is a diagnostic, not a substitute for evaluating the integral: a result below the projected area signals an error in the setup or arithmetic.
Check Your Understanding
For each question, identify the graph-area factor and the appropriate domain bounds before attempting the evaluation.
- What is the graph-area factor for \(z=3x-4y\) over a rectangle?
- Write the polar-coordinate surface-area integral for \(z=x^2+y^2\) over a disk of radius \(R\).
- For \(z=xy\), why does the graph-area factor over a centered disk depend only on \(r\), despite \(xy\) not being radial?
- Write the polar bounds for an annulus with inner radius \(2\) and outer radius \(5\).
- What factor must accompany \(dr\,d\theta\) when transforming \(dA\) to polar coordinates, and why?