Chapter 20

Scalar Surface Integrals

We've learned so far to integrate multivariate functions over curves and over regions in $ℝ 2$ or $ℝ 3$ . Here we extend our knowledge to consider integrals over a surface in space. Such integrals appear in many mathematical and physical contexts, and in particular turn out to be rather important in our final unit, dealing with vector fields.

20.1Parameterized Surfaces and Surface Area

Several chapters ago, we learned to integrate a scalar function along a curve. The construction for a surface is the same idea one dimension up:

| Curve | Surface | |---|---| | $⃗ 𝑐 (𝑡)$ | $⃗ 𝑟 (𝑢, 𝑣)$ | | small interval $𝑑 𝑡$ | small rectangle $𝑑 𝑢 𝑑 𝑣$ | | tangent vector $⃗ 𝑐' (𝑡)$ | tangent vectors $⃗ 𝑟 𝑢, ⃗ 𝑟 𝑣$ | | length scale $‖ ⃗ 𝑐' (𝑡) ‖$ | area scale $‖ ⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣 ‖$ | | $𝑑 𝑠 = ‖ ⃗ 𝑐' (𝑡) ‖ 𝑑 𝑡$ | $𝑑 𝑆 = ‖ ⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣 ‖ 𝑑 𝑢 𝑑 𝑣$ |

Recall the complete scalar line-integral formula:

\int 𝐶 𝑓 𝑑 𝑠 = \int 𝑏 𝑎 𝑓 (⃗ 𝑐 (𝑡)) ‖ ⃗ 𝑐' (𝑡) ‖ 𝑑 𝑡 .

We plug the curve into the scalar field, multiply by the geometric length scale, and integrate over the parameter interval. For a surface, a two-dimensional parameter domain replaces that interval.

Definition 20.1 (Parameterized Surface). A parameterized surface is the image of a map

⃗ 𝑟 : 𝐷 \subseteq ℝ 2 ⟶ ℝ 3, ⃗ 𝑟 (𝑢, 𝑣) = ⟨ 𝑥 (𝑢, 𝑣), 𝑦 (𝑢, 𝑣), 𝑧 (𝑢, 𝑣) ⟩ .

The region $𝐷$ is called the parameter domain.

This is the same kind of map we studied in the previous chapter, except that its image now lies in three-dimensional space instead of another plane.

To get an infinitesimal surface area element, we follow the same trick that worked for graphs: if we can find two tangent vectors at a point, we can find the area of the infinitesimal parallelogram they span using the cross product.

Here the tangent vectors are the $𝑢$ and $𝑣$ derivatives of the parameterization:

⃗ 𝑟 𝑢 = 𝜕 ⃗ 𝑟 𝜕 𝑢 = ⟨ 𝜕 𝑥 𝜕 𝑢, 𝜕 𝑦 𝜕 𝑢, 𝜕 𝑧 𝜕 𝑢 ⟩,

⃗ 𝑟 𝑣 = 𝜕 ⃗ 𝑟 𝜕 𝑣 = ⟨ 𝜕 𝑥 𝜕 𝑣, 𝜕 𝑦 𝜕 𝑣, 𝜕 𝑧 𝜕 𝑣 ⟩ .

A small rectangle with side lengths $𝑑 𝑢$ and $𝑑 𝑣$ maps approximately to the parallelogram spanned by

⃗ 𝑟 𝑢 𝑑 𝑢 a n d ⃗ 𝑟 𝑣 𝑑 𝑣 .

The cross product $⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣$ is normal to the surface, and its magnitude gives the area scale:

𝑑 𝑆 = ‖ ⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣 ‖ 𝑑 𝑢 𝑑 𝑣 .

Definition 20.2 (Surface Area from a Parameterization). If $⃗ 𝑟 (𝑢, 𝑣)$ parameterizes a surface $𝑆$ once over a parameter domain $𝐷$ , then

A r e a (𝑆) = \iint 𝐷 ‖ ⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣 ‖ 𝑑 𝑢 𝑑 𝑣 .

In practice, we require $⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣 \neq ⃗ 0$ in the regular interior of the parameter domain. We also want the parameterization to cover the surface once, apart from harmless identifications along the boundary. If a parameterization covers a patch twice, the integral counts its area twice.

20.1.1Graphs of Functions

The most important computational case in this course is a surface that is the graph of a function $𝑧 = 𝑔 (𝑥, 𝑦)$ . We can parameterize the graph by

⃗ 𝑟 (𝑥, 𝑦) = ⟨ 𝑥, 𝑦, 𝑔 (𝑥, 𝑦) ⟩ .

Taking the partial derivatives gives a tangent vector in the $𝑥$ direction and one in the $𝑦$ direction:

⃗ 𝑟 𝑥 = 𝜕 𝜕 𝑥 ⟨ 𝑥, 𝑦, 𝑔 (𝑥, 𝑦) ⟩ = ⟨ 1, 0, 𝑔 𝑥 ⟩,

⃗ 𝑟 𝑦 = 𝜕 𝜕 𝑦 ⟨ 𝑥, 𝑦, 𝑔 (𝑥, 𝑦) ⟩ = ⟨ 0, 1, 𝑔 𝑦 ⟩ .

These two vectors span an infinitesimal parallelogram, whose area is the area element we seek. How do we find the area of a small parallelogram again? The cross product! Thus,

⃗ 𝑟 𝑥 \times ⃗ 𝑟 𝑦 = ∣ ∣ ∣ ∣ ⃗ 𝑖 ⃗ 𝑗 ⃗ 𝑘 10 𝑔 𝑥 01 𝑔 𝑦 ∣ ∣ ∣ ∣ = ⟨ - 𝑔 𝑥, - 𝑔 𝑦, 1 ⟩,

and

‖ ⃗ 𝑟 𝑥 \times ⃗ 𝑟 𝑦 ‖ = \sqrt 1 + 𝑔 2 𝑥 + 𝑔 2 𝑦 .

Definition 20.3 (Area Element on a Graph). The area element along the graph of a function $𝑧 = 𝑔 (𝑥, 𝑦)$ is

𝑑 𝑆 = \sqrt 1 + 𝑔 2 𝑥 + 𝑔 2 𝑦 𝑑 𝑥 𝑑 𝑦 .

Example 20.4 (Area of a Paraboloid). Find the area of the paraboloid $𝑧 = 𝑥 2 + 𝑦 2$ that lies below the plane $𝑧 = 9$ .

Here

𝑔 𝑥 = 2 𝑥, 𝑔 𝑦 = 2 𝑦,

𝑑 𝑆 = \sqrt 1 + 4 𝑥 2 + 4 𝑦 2 𝑑 𝑥 𝑑 𝑦 .

The paraboloid meets the plane when

𝑥 2 + 𝑦 2 = 9,

so its projection onto the $𝑥 𝑦$ -plane is the disc $𝑥 2 + 𝑦 2 \leq 9$ . Converting that double integral to polar coordinates gives

A r e a (𝑆) = \iint 𝑥 2 + 𝑦 2 \leq 9 \sqrt 1 + 4 𝑥 2 + 4 𝑦 2 𝑑 𝑥 𝑑 𝑦 = \int 2 𝜋 0 \int 30 \sqrt 1 + 4 𝑟 2 𝑟 𝑑 𝑟 𝑑 𝜃 = 2 𝜋 [112 (1 + 4 𝑟 2) 3 / 2] 30 = 𝜋 6 (37 \sqrt 37 - 1) .

20.1.2Cylinders and Spheres

The cylinder and sphere formulas from the coordinate-systems chapter are quick checks of this general construction. A cylinder of radius $𝑅$ can be parameterized by

⃗ 𝑟 (𝜃, 𝑧) = ⟨ 𝑅 c o s 𝜃, 𝑅 s i n 𝜃, 𝑧 ⟩ .

Its two tangent vectors have cross-product magnitude $𝑅$ , so

𝑑 𝑆 = 𝑅 𝑑 𝜃 𝑑 𝑧 .

A sphere of radius $𝑅$ can be parameterized by

⃗ 𝑟 (𝜃, 𝜙) = ⟨ 𝑅 c o s 𝜃 s i n 𝜙, 𝑅 s i n 𝜃 s i n 𝜙, 𝑅 c o s 𝜙 ⟩ .

Here the cross-product magnitude is $𝑅 2 s i n 𝜙$ , giving

𝑑 𝑆 = 𝑅 2 s i n 𝜙 𝑑 𝜙 𝑑 𝜃 .

These are not separate tricks: they are both instances of $𝑑 𝑆 = ‖ ⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣 ‖ 𝑑 𝑢 𝑑 𝑣$ .

20.2Scalar Surface Integrals

We are now interested in integrating over a surface instead of a curve. If that surface $𝑆$ is a 2D region inside of the $𝑥 𝑦$ -plane, we already know how to do this: it is just the double integral $\iint 𝑆 𝑓 𝑑 𝐴$ , where $𝑑 𝐴$ is an infinitesimal piece of area on the plane. So, what we are interested in here instead is when $𝑆$ is a surface in 3-dimensional space and $𝑓 (𝑥, 𝑦, 𝑧)$ is a scalar function on 3-dimensional space.

These integrals add up the total amount of $𝑓$ which lies on the surface $𝑆$ . This is useful for many things: if $𝑓$ is a surface density, this gives the mass of the object. If $𝑓$ is a charge density, this gives the total charge on the surface---a computation that is useful inside batteries, where the anode and cathode may be tightly coiled surfaces. These integrals can also be used to help find the average value of a function: divide the surface integral of $𝑓$ over $𝑆$ by the surface area of $𝑆$ .

A scalar field does not stop existing away from the surface; its values there simply do not contribute. We first restrict the field to $𝑆$ , then accumulate only those values against surface area.

Figure 20.1 A scalar field doesn't stop existing just because we've decided to integrate over a surface instead of a solid. On the left it fills the room exactly as it did for a triple integral — warm where the haze is red, cool where it's blue — and the pale sheet sitting inside it is the surface

𝑆

we're about to integrate over. Pull that same sheet out on the right and paint it by the very field it was sitting in: same colours, same values, just read off two dimensions instead of three. Everywhere else in the room,

𝑓

still doesn't matter here — a surface integral only ever sees

𝑓

restricted to

𝑆

We denote such an integral by

\iint 𝑆 𝑓 𝑑 𝑆,

where we think of $𝑑 𝑆$ as an infinitesimal piece of surface area. A lot of the work in computing surface integrals is finding the right form for $𝑑 𝑆$ , so that we can convert the surface integral into a double integral we know how to do.

Definition 20.5 (Scalar Surface Integral). If $⃗ 𝑟 (𝑢, 𝑣)$ parameterizes a surface $𝑆$ once over a parameter domain $𝐷$ , then

\iint 𝑆 𝑓 𝑑 𝑆 = \iint 𝐷 𝑓 (⃗ 𝑟 (𝑢, 𝑣)) ‖ ⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣 ‖ 𝑑 𝑢 𝑑 𝑣 .

This definition has two parts, exactly as the scalar line-integral formula did:

p u l l t h e s c a l a r f i e l d b a c k t o t h e p a r a m e t e r d o m a i n + i n c l u d e t h e g e o m e t r i c a r e a s c a l e .

The first factor is

𝑓 (⃗ 𝑟 (𝑢, 𝑣)) = 𝑓 (𝑥 (𝑢, 𝑣), 𝑦 (𝑢, 𝑣), 𝑧 (𝑢, 𝑣)),

and the second is

‖ ⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣 ‖ .

The simplest scalar surface integral takes $𝑓 = 1$ and merely adds up the pieces of area. A surface density $𝜌$ weights each area element to give mass, and dividing by total area gives an average value:

A r e a (𝑆) = \iint 𝑆 1 𝑑 𝑆, 𝑀 = \iint 𝑆 𝜌 𝑑 𝑆, 𝑓 a v g = 1 A r e a (𝑆) \iint 𝑆 𝑓 𝑑 𝑆 .

The word density must be read with its units. A line density has units of mass per length and is integrated against $𝑑 𝑠$ ; a surface density has units of mass per area and is integrated against $𝑑 𝑆$ ; a volume density has units of mass per volume and is integrated against $𝑑 𝑉$ .

For a graph $𝑧 = 𝑔 (𝑥, 𝑦)$ over a region $𝑅$ in the $𝑥 𝑦$ -plane, the general definition becomes the practical formula

\iint 𝑆 𝑓 𝑑 𝑆 = \iint 𝑅 𝑓 (𝑥, 𝑦, 𝑔 (𝑥, 𝑦)) \sqrt 1 + 𝑔 2 𝑥 + 𝑔 2 𝑦 𝑑 𝑥 𝑑 𝑦 .

We evaluate $𝑓$ at points along the graph and integrate with respect to its surface-area element.

Example 20.6 (A Scalar Integral over a Graph). Find $\iint 𝑆 𝑦 𝑑 𝑆$ for $𝑆$ the surface $𝑧 = 𝑥 + 𝑦 2$ over the rectangle $0 \leq 𝑥 \leq 1$ , $0 \leq 𝑦 \leq 2$ .

The scalar field is already $𝑓 (𝑥, 𝑦, 𝑧) = 𝑦$ , so restricting it to the graph does not change its formula. For $𝑔 (𝑥, 𝑦) = 𝑥 + 𝑦 2$ ,

𝑔 𝑥 = 1, 𝑔 𝑦 = 2 𝑦,

and hence

𝑑 𝑆 = \sqrt 1 + 12 + (2 𝑦) 2 𝑑 𝑥 𝑑 𝑦 = \sqrt 2 + 4 𝑦 2 𝑑 𝑥 𝑑 𝑦 .

Therefore

\iint 𝑆 𝑦 𝑑 𝑆 = \int 20 \int 10 𝑦 \sqrt 2 + 4 𝑦 2 𝑑 𝑥 𝑑 𝑦 = \int 20 𝑦 \sqrt 2 + 4 𝑦 2 𝑑 𝑦 = 112 [(2 + 4 𝑦 2) 3 / 2] 20 = 13 \sqrt 2 3 .

20.2.1Orientation and Additivity

Scalar surface integrals are orientation-independent. Swapping the two parameters reverses the cross product,

⃗ 𝑟 𝑣 \times ⃗ 𝑟 𝑢 = - (⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣),

but it does not change its magnitude:

‖ ⃗ 𝑟 𝑣 \times ⃗ 𝑟 𝑢 ‖ = ‖ ⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣 ‖ .

Thus a scalar area element has no preferred direction. Later, when we measure the flux of a vector field through a surface, we will replace the scalar area $𝑑 𝑆$ with an oriented vector area $𝑑 ⃗ 𝑆 = ⃗ 𝑛 𝑑 𝑆$ . Orientation will matter there.

Just like for line integrals, surface integrals can be computed over surfaces made from two or more pieces by computing each part separately and adding the results. If $𝑆 1$ and $𝑆 2$ have disjoint interiors and meet only along boundary curves, then

\iint 𝑆 1 \cup 𝑆 2 𝑓 𝑑 𝑆 = \iint 𝑆 1 𝑓 𝑑 𝑆 + \iint 𝑆 2 𝑓 𝑑 𝑆 .

20.2.2Further Practice

The following legacy examples are useful applications of the formulas above, but they do not require new theory.

Exercise 20.7. Compute $\iint 𝑆 𝑥 3 𝑑 𝑆$ for $𝑆$ the cylinder of radius $2$ centered on the $𝑧$ -axis between $𝑧 = 1$ and $𝑧 = 3$ .

Exercise 20.8. Compute $\iint 𝑆 𝑥 𝑑 𝑆$ for $𝑆$ the cylinder of radius $1$ centered on the $𝑥$ -axis from $𝑥 = 0$ to $𝑥 = 2$ .

Exercise 20.9. Compute $\iint 𝑆 𝑧 𝑑 𝑆$ for $𝑆$ the upper hemisphere of the sphere of radius $2$ .

Exercise 20.10. Compute $\iint 𝑆 (𝑥 2 + 𝑦 2 + 𝑧 2) 𝑑 𝑆$ for $𝑆$ the sphere of radius $3$ centered at the origin.

Exercise 20.11. Let $𝑆$ denote the closed surface formed by the paraboloid $𝑧 = 2 - 𝑥 2 - 𝑦 2$ and the plane $𝑧 = 1$ . Set up $\iint 𝑆 (𝑥 + 𝑧) 𝑑 𝑆$ by splitting $𝑆$ into its two pieces.