Download original notes
PAGE 1

17.6 Surface Integrals (part 1)

line integral: \(\int_C f(x, y, z) ds\)

accumulation of \(f(x, y, z)\) along a curve \(C\) parametrized

A smooth curve labeled C in space with a small segment ds highlighted by arrows.

surface integral: accumulation of \(f(x, y, z)\) all over a surface \(S\) (capital S)

A 3D coordinate system showing a surface S with a small patch labeled ds.

tiny patch of the surface \(S\)

call this \(dS\)

(capital S)

the total accumulation over this surface looks like

\[ \iint_S f(x, y, z) dS \]

PAGE 2

to compute a line integral \(\int_C f ds\), we need to parametrize the curve \(C\)

example: line from \((-1, 1)\) to \((2, 4)\)

\[ \vec{r}(t) = \langle -1, 1 \rangle + t \langle 3, 3 \rangle = \langle -1 + 3t, 1 + 3t \rangle \]

\(0 \le t \le 1\)

parametrization is not unique

to compute a surface integral \(\iint_S f dS\), we need to parametrize \(S\)

this is one dimension higher than curve, so one more parameter

\[ \text{surface: } \vec{r}(u, v) \]

then the domains of \(u, v\)

parameters
PAGE 3

Example: Parametrizing a Plane in the First Octant

Parametrize the part of \(3x + 2y + z = 6\) in the first octant.

3D coordinate system showing a plane intersecting axes at (2,0,0), (0,3,0), and (0,0,6) in the first octant.

Figure 1: Visualization of the plane in the first octant.

parametrize: how to locate points on the surface

one possible way: specify \(x, y\), then \(z = 6 - 3x - 2y\)

so, we can let \(u = x, v = y\), then

\[ z = f(x, y) = f(u, v) = 6 - 3u - 2v \]

One Possible Parametrization

\[ \vec{r}(u, v) = \langle u, v, 6 - 3u - 2v \rangle \]

Where \(u\) corresponds to \(x\) and \(v\) corresponds to \(y\).

Determining Bounds for \(u, v\)

2D graph in the xy-plane showing a line from (0,3) to (2,0) representing the boundary of the region.

Figure 2: Projection onto the xy-plane to find bounds.

From the projection:

  • \(y = 3 - \frac{3}{2}x\)
  • \(v = 3 - \frac{3}{2}u\)

Bounds:

\[ 0 \le u \le 2, \quad 0 \le v \le 3 - \frac{3}{2}u \]

Final Parametrization:

\[ \vec{r}(u, v) = \langle u, v, 6 - 3u - 2v \rangle \]\[ 0 \le u \le 2, \quad 0 \le v \le 3 - \frac{3}{2}u \]
PAGE 4

Example: Parametrizing a Sphere Above a Plane

\(x^2 + y^2 + z^2 = 9\) above \(z = 1\) in the first octant.

3D sketch of a sphere portion in the first octant above the plane z=1, showing a circular boundary.

Figure 3: Sphere portion in the first octant.

One choice of parametrization

\(x = u, y = v, z = \sqrt{9 - x^2 - y^2} = \sqrt{9 - u^2 - v^2}\)

\[ \vec{r}(u, v) = \langle u, v, \sqrt{9 - u^2 - v^2} \rangle \]

Finding the Bounds

At \(z = 1 \rightarrow x^2 + y^2 = 8\), which is a circle of radius \(\sqrt{8}\).

2D graph of a quarter circle in the first quadrant with radius square root of 8.

Figure 4: Projection of the boundary onto the xy-plane.

Boundary equations:

  • \(y = \sqrt{8 - x^2}\)
  • \(v = \sqrt{8 - u^2}\)

Bounds for parameters:

\[ 0 \le u \le \sqrt{8}, \quad 0 \le v \le \sqrt{8 - u^2} \]

Final Parametrization:

\[ \vec{r}(u, v) = \langle u, v, \sqrt{9 - u^2 - v^2} \rangle \]\[ 0 \le u \le \sqrt{8}, \quad 0 \le v \le \sqrt{8 - u^2} \]
PAGE 5

Parametrization in Cylindrical Coordinates

Can we do this in another coord. system, for example, cylindrical?

3D plot of a spherical surface segment in the first octant with axes x, y, and z.
2D projection on the xy-plane showing a quarter circle of radius sqrt(8) in the first quadrant.

onto xy-plane

\[ 0 \le r \le \sqrt{8} \]\[ 0 \le \theta \le \frac{\pi}{2} \]\[ 1 \le z \le \sqrt{9-r^2} \]

choose two to be parameters: \( u = r, v = \theta \)

then

\[ x = r \cos \theta = u \cos v \]\[ y = r \sin \theta = u \sin v \]
\[ 0 \le u \le \sqrt{8} \]\[ 0 \le v \le \frac{\pi}{2} \]
\[ \vec{r}(u,v) = \langle u \cos v, u \sin v, \sqrt{9-u^2} \rangle \]
\( x = r \cos \theta \)
\( y = r \sin \theta \)
\( z = z \)
\[ 0 \le u \le \sqrt{8}, \quad 0 \le v \le \frac{\pi}{2} \]
PAGE 6

Surface Integrals and Area Elements

\[ \int_C f \, ds \]
line integral \( ds = |\vec{r}'| \, dt \)
length of small segment
\[ \iint_S f \, dS \]
surf. integral \( dS = ? \)
area of small patch of surface
3D plot showing a surface S in xyz-space with a differential area element dS highlighted.
Detailed view of a surface patch as a parallelogram defined by vectors a and b at point r(u,v).

area of \( dS \) is \( |\vec{a} \times \vec{b}| \)

recall:

\[ \frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x,y)}{h} = f_x \]

so, \( f_x \approx \frac{f(x+\Delta x, y) - f(x,y)}{h} \)

or \( f(x+\Delta x, y) - f(x,y) \approx f_x h \)

Geometric diagram of a parallelogram with sides a and b, labeled with area equals magnitude of a cross b.
PAGE 7

now using the above, we got

\[ \vec{a} = \vec{r}(u, v+\Delta v) - \vec{r}(u,v) \approx \vec{r}_v \Delta v \]\[ \vec{b} = \vec{r}(u+\Delta u, v) - \vec{r}(u,v) \approx \vec{r}_u \Delta u \]\[ dS = |\vec{a} \times \vec{b}| = |\vec{r}_v \Delta v \times \vec{r}_u \Delta u| \]\[ = |\vec{r}_u \times \vec{r}_v| \Delta u \Delta v \]

\( \vec{r}_v = \frac{\partial}{\partial v}(\vec{r}(u,v)) \)

now let \( \Delta u, \Delta v \to du, dv \)

so, \( dS = |\vec{r}_u \times \vec{r}_v| du dv \)

this goes in \( \iint_S f dS \)

PAGE 8

Example

\( \iint_S (x+y) dS \)

\( S \): part of plane \( 3x + 2y + z = 6 \) in the first octant above \( 0 \le x \le 1, 0 \le y \le 3/2 \)

3D coordinate graph showing a plane segment in the first octant bounded by x=1 and y=1.5.

let's reuse parametrization from earlier: \( u=x, v=y, z=6-3u-2v \)

\( \vec{r}(u,v) = \langle u, v, 6-3u-2v \rangle \)

\( 0 \le u \le 1, \quad 0 \le v \le 3/2 \)

now, \( dS = |\vec{r}_u \times \vec{r}_v| du dv \)

\( dS = \sqrt{14} du dv \)

\( \vec{r}_u = \langle 1, 0, -3 \rangle \)

\( \vec{r}_v = \langle 0, 1, -2 \rangle \)

\( \vec{r}_u \times \vec{r}_v = \langle 3, 2, 1 \rangle \)

\( |\vec{r}_u \times \vec{r}_v| = \sqrt{14} \)

PAGE 9

Surface Integral Calculation

\[ \iint_{S} (x+y) \, dS \]

Where the components are defined as:

  • \( dS \rightarrow \sqrt{14} \, du \, dv \)
  • \( x \)-component of \( \vec{r}(u,v) \rightarrow u \)
  • \( y \)-component of \( \vec{r}(u,v) \rightarrow v \)

Bounds:

\( 0 \le u \le 1 \)

\( 0 \le v \le 3/2 \)

Here, any order is ok since both are constant-bounded.

Substituting the components and bounds into the integral:

\[ = \int_{0}^{3/2} \int_{0}^{1} (u+v) \sqrt{14} \, du \, dv \]

Note: The inner integral is with respect to \( u \) and the outer integral is with respect to \( v \).

\( = \dots = \)
\[ \frac{15\sqrt{14}}{8} \]

Possible Interpretation:

If \( (x+y) \) is the density of a material, then this integral represents the mass.