How much paint covers a curved roof? The partial derivatives tell how much a surface tilts, and tilting stretches area: a patch of the surface is larger than its shadow on the floor. A double integral adds up the stretched patches.
Key idea
The area of the part of $z = f(x, y)$ above a region $D$ is
$$A = \iint_D \sqrt{1 + f_x^2 + f_y^2}\,dA.$$
The factor $\sqrt{1 + f_x^2 + f_y^2}$ is how much a small patch of $D$ is stretched when lifted onto the surface.
Why it works
Over a small rectangle $\Delta x$ by $\Delta y$, the surface is nearly its tangent plane, a parallelogram with sides $\langle \Delta x, 0, f_x\Delta x \rangle$ and $\langle 0, \Delta y, f_y\Delta y \rangle$ (step in $x$, rise $f_x\Delta x$; step in $y$, rise $f_y\Delta y$). Its area is the length of the cross product:
$$\left|\langle -f_x, -f_y, 1 \rangle\right|\Delta x\,\Delta y = \sqrt{1 + f_x^2 + f_y^2}\;\Delta x\,\Delta y.$$
Adding the parallelograms and refining gives the integral. A flat patch ($f_x = f_y = 0$) is not stretched at all.
Worked examples
Worked example: A plane
Find the area of the part of the plane $z = 5 - 2x - 2y$ above the rectangle ${}0 \le x \le 2$, ${}0 \le y \le 3$.
- $f_x = -2$, $f_y = -2$: the factor is $\sqrt{1 + 4 + 4} = 3$ everywhere.
- Area $$= 3 \times (\text{area of the rectangle}) = 3 \cdot 6 = 18.$$
Answer: ${}18$
Worked example: A paraboloid
Find the area of the part of $z = x^2 + y^2$ below the plane $z = 2$.
- The shadow is the disk $x^2 + y^2 \le 2$. The factor is $\sqrt{1 + 4x^2 + 4y^2} = \sqrt{1 + 4r^2}$.
- In polar coordinates: $A = \int_0^{2\pi}\int_0^{\sqrt{2}} \sqrt{1 + 4r^2}\,r\,dr\,d\theta$.
- With $u = 1 + 4r^2$: $$\int_0^{\sqrt{2}} \sqrt{1 + 4r^2}\,r\,dr = \frac{1}{12}\left[(1 + 4r^2)^{3/2}\right]_0^{\sqrt{2}} = \frac{1}{12}(27 - 1) = \frac{13}{6}.$$
- $A = 2\pi \cdot \frac{13}{6} = \frac{13\pi}{3}$.
Answer: $\frac{13\pi}{3}$
Worked example: A cone
Find the area of the cone $z = \sqrt{x^2 + y^2}$ above the disk $x^2 + y^2 \le 9$.
- $f_x = \frac{x}{\sqrt{x^2 + y^2}}$, $f_y = \frac{y}{\sqrt{x^2 + y^2}}$, so ${}1 + f_x^2 + f_y^2 = 2$.
- Area $= \sqrt{2} \cdot 9\pi = 9\sqrt{2}\,\pi$.
Answer: ${}9\sqrt{2}\,\pi$
Common mistake
Integrating the shadow instead of the surface. $\iint_D dA$ is the area of the region on the floor, which is always smaller than (or equal to) the surface above it. The factor $\sqrt{1 + f_x^2 + f_y^2} \ge 1$ must be inside the integral, and the ${}1$ under the root must not be dropped. Check: for a tilted plane the answer is a constant multiple, bigger than 1, of the shadow's area.