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17.6 Surface Integrals (part 2)

Last time:

\[ \iint_{S} g(x, y, z) \, dS = \iint_{S} g(u, v) |\vec{r}_u \times \vec{r}_v| \, dA \]

General formula, works in any coordinate system. Note: \( dA \) can be \( du \, dv \) or \( dv \, du \).

Often we use Cartesian: \( u = x \), \( v = y \). The resulting formula is useful to know.

Surface \( S \): \( \vec{r}(u, v) = \vec{r}(x, y) = \langle x, y, f(x, y) \rangle \) (where \( z = f(x, y) \))

\[ \vec{r}_u = \vec{r}_x = \langle 1, 0, f_x \rangle \]\[ \vec{r}_v = \vec{r}_y = \langle 0, 1, f_y \rangle \]
\[ \vec{r}_u \times \vec{r}_v = \langle -f_x, -f_y, 1 \rangle \]\[ |\vec{r}_u \times \vec{r}_v| = \sqrt{1 + f_x^2 + f_y^2} \]
\[ dS = \sqrt{1 + f_x^2 + f_y^2} \, dy \, dx \quad \text{or} \quad dx \, dy \]

\( \rightarrow \) area of a small patch of \( S \)

Notice similarity to length of curve \( L = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2} \, dx \) if \( y = f(x) \)

So, in Cartesian, the surface integral can be written as

\[ \iint_{R} g(x, y, z) \sqrt{1 + f_x^2 + f_y^2} \, dA \]
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Example

Evaluate:

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

\( S \): \( z = 6 - 3x - 2y \) above the rectangle \( 0 \le x \le 1 \), \( 0 \le y \le 3/2 \)

3D coordinate system showing a plane segment in the first octant with intercepts at z=6, x=2, and y=3.

Surface: \( z = f(x, y) = 6 - 3x - 2y \)

\( f_x = -3 \), \( f_y = -2 \)

\[ dS = \sqrt{1 + f_x^2 + f_y^2} \, dA = \sqrt{14} \, dy \, dx \]
\[ \iint_{S} (x + y) \, dS = \int_{0}^{1} \int_{0}^{3/2} (x + y) \sqrt{14} \, dy \, dx \]
\[ = \dots = \frac{15\sqrt{14}}{8} \]
2D xy-plane showing a rectangular region bounded by x=1 and y=3/2 within a triangle bounded by y=3-3/2x.
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Surface Area Example

Example: Find the surface area of \( z^2 = x^2 + y^2 \) for \( 0 \le z \le 2 \).

We can rewrite the surface as a function of \( x \) and \( y \):

\[ z = f(x,y) = \sqrt{x^2 + y^2} \]
A 3D coordinate system showing a cone opening upwards from the origin, capped at z=2.

Projection onto xy-plane

The projection of the cone \( z = \sqrt{x^2 + y^2} \) with \( z \le 2 \) onto the \( xy \)-plane is a disk of radius 2.

The boundaries are given by:

  • \( y = \sqrt{4 - x^2} \) (upper semi-circle)
  • \( y = -\sqrt{4 - x^2} \) (lower semi-circle)

The region is defined by:

\[ -2 \le x \le 2 \quad \text{and} \quad -\sqrt{4 - x^2} \le y \le \sqrt{4 - x^2} \]
A 2D circle of radius 2 centered at the origin in the xy-plane, labeled with its upper and lower boundary equations.

Surface Area Differential

The differential surface area element is:

\[ dS = \sqrt{1 + f_x^2 + f_y^2} \, dA \]

Calculating partial derivatives:

\[ f_x = \frac{x}{\sqrt{x^2 + y^2}} \quad f_y = \frac{y}{\sqrt{x^2 + y^2}} \]

Substituting into the formula:

\[ dS = \sqrt{1 + \frac{x^2}{(\sqrt{x^2 + y^2})^2} + \frac{y^2}{(\sqrt{x^2 + y^2})^2}} = \sqrt{\frac{2(x^2 + y^2)}{x^2 + y^2}} = \sqrt{2} \]

Thus,

\[ dS = \sqrt{2} \, dy \, dx \]
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Surface Area Calculation

\[ \text{Surface area:} \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \sqrt{2} \, dy \, dx \]
\[ = \sqrt{2} \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} dy \, dx \]

Geometric Interpretation:

The double integral represents the area of a circle with radius 2.

\[ \text{Area} = \pi(2)^2 = 4\pi \]

A 2D circle of radius 2 in the xy-plane, illustrating the region of integration.

Therefore,

\[ = \sqrt{2} \cdot 4\pi = 4\sqrt{2}\pi \]

Alternative: Change to Polar Coordinates

Using the fact that \( dA = r \, dr \, d\theta \) in polar coordinates:

\[ \sqrt{2} \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} dy \, dx = \sqrt{2} \int_{0}^{2\pi} \int_{0}^{2} r \, dr \, d\theta \]

Evaluating this integral yields the same result:

\[ = 4\sqrt{2}\pi \]
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\[ dS = \sqrt{1 + f_x^2 + f_y^2} \, dA \text{ is assumed to be in Cartesian} \]

if we change to anything else, we need to manually adjust \( dA \)

but \( dS = |\vec{r}_u \times \vec{r}_v| \, dA \) ALWAYS accounts for \( dA \) automatically

Example (same cone)

\[ z^2 = x^2 + y^2 \rightarrow z = \sqrt{x^2 + y^2} \rightarrow z = \sqrt{r^2} = r \] \[ 0 \le z \le 2 \]

now let's parametrize in cylindrical

A circle centered at the origin in the xy-plane with radius 2.
\[ 0 \le r \le 2 \] \[ 0 \le \theta \le 2\pi \] \[ \downarrow \] \[ 0 \le u \le 2 \] \[ 0 \le v \le 2\pi \]
\[ u = r, \quad v = \theta \] \[ x = r \cos \theta = u \cos v \] \[ y = r \sin \theta = u \sin v \] \[ z = z = r = u \]
\[ \vec{r}(u,v) = \langle u \cos v, u \sin v, u \rangle \] \[ \vec{r}_u = \langle \cos v, \sin v, 1 \rangle \] \[ \vec{r}_v = \langle -u \sin v, u \cos v, 0 \rangle \] \[ \vec{r}_u \times \vec{r}_v = \langle -u \cos v, -u \sin v, u \rangle \] \[ |\vec{r}_u \times \vec{r}_v| = \sqrt{2} \, u \]
\[ dS = |\vec{r}_u \times \vec{r}_v| \, dA = \sqrt{2} \, u \, du \, dv \]

notice \( |\vec{r}_u \times \vec{r}_v| \) ALWAYS provides the "missing" piece in \( dA \) (e.g., \( r \, dr \, d\theta \))

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we can calculate the average value of \( f(x,y,z) \) on surface \( S \)

\[ f_{\text{avg}} = \frac{\iint_S f \, dS}{\iint_S dS} \]

physically meaning:

\[ \iint_S f \, dS = \text{mass if } f \text{ is density} \]
\[ f_{\text{avg}} \underbrace{\iint_S dS}_{\text{Surface area}} = \text{mass with constant density } f_{\text{avg}} \text{ over the same surface} \]
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Example: Average Distance on a Surface

Find the average distance of points on \( z = x^2 + y^2 \), \( 1 \le z \le 4 \) from the origin.

A 3D paraboloid surface z = x squared plus y squared between z equals 1 and z equals 4.

\( d = \sqrt{x^2 + y^2 + z^2} \) distance from \( (0, 0, 0) \)

This is the function to average.

Parametrization

Coord. sys? Cylindrical is good.

Top-down view of concentric circles in the xy-plane representing the bounds of the paraboloid.
  • At \( z = 4 \rightarrow x^2 + y^2 = 4 \rightarrow r^2 = 4 \rightarrow r = 2 \)
  • At \( z = 1 \rightarrow r = 1 \)

\( 1 \le r \le 2 \)

\( 0 \le \theta \le 2\pi \)

\( u = r \)

\( v = \theta \)

\( 1 \le u \le 2 \)

\( 0 \le v \le 2\pi \)

\( \vec{r}(u, v) = \langle u \cos v, u \sin v, u^2 \rangle \)

Note: \( z = x^2 + y^2 = r^2 = u^2 \)

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Partial Derivatives and Surface Area Element

\( \vec{r}_u = \langle \cos v, \sin v, 2u \rangle \)

\( \vec{r}_v = \langle -u \sin v, u \cos v, 0 \rangle \)

\( |\vec{r}_u \times \vec{r}_v| = \dots = u \sqrt{1 + 4u^2} \)

Average Value Formula

\( f_{\text{avg}} = \frac{\iint_S f \, dS}{\iint_S dS} \)

Calculations

Numerator (Integral of distance):

\( \iint_S f \, dS = \int_0^{2\pi} \int_1^2 \sqrt{u^2 + u^4} \cdot u \sqrt{1 + 4u^2} \, du \, dv = \dots = 2\pi (15.21) \)

Where \( f = d = \sqrt{x^2 + y^2 + z^2} = \sqrt{u^2 + u^4} \) and \( dS = u \sqrt{1 + 4u^2} \, du \, dv \)

Denominator (Surface Area):

\( \iint_S dS = \int_0^{2\pi} \int_1^2 u \sqrt{1 + 4u^2} \, du \, dv = \dots = 2\pi (4.91) \)

Final Result

\( f_{\text{avg}} = \frac{2\pi (15.21)}{2\pi (4.91)} \approx 3.1 \)