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17.6 Surface Integrals (part 3)

Last two times: surface integrals in scalar fields

\[ \iint_S f(x,y,z) dS = \iint_R f(u,v) |\vec{r}_u \times \vec{r}_v| dA \]

area of small patch of surface S

Consider a surface in the 3D coordinate system with parameters \(u\) and \(v\). A small patch on the surface is bounded by curves of constant \(u\) and constant \(v\).

Figure: 3D axes showing a surface patch with tangent vectors r_u du and r_v dv forming a parallelogram area.

Area of the patch:

\[ \text{area} = |\vec{r}_u \times \vec{r}_v| du dv \]

But \(\vec{r}_u \times \vec{r}_v\) is also the normal vector (so is \(\vec{r}_v \times \vec{r}_u\)).

Figure: Diagram of a parallelogram with two opposite normal vectors: r_u cross r_v and r_v cross r_u.

In a scalar field, we work with the magnitude of the normal, so the order of the cross product doesn't matter.
In a vector field, direction matters. So, which one to choose?

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By convention, we choose the normal vector that is pointing upward or outward unless otherwise specified.

A surface with the "correct" normal is said to be positively-oriented.

Surface integral in vector field:

\[ \iint_S \vec{F} \cdot d\vec{S} \]

\(\vec{F}\): vector field
\(d\vec{S}\): denotes an oriented surface

\[ = \iint_S \vec{F} \cdot \vec{n} dS \]

\(dS\): same \(dS\) as in scalar field version
\(\vec{n}\): unit normal in the correct direction

This integral accumulates the component of \(\vec{F}\) that is in the same direction as \(\vec{n}\). This integral is also called the Flux integral.

Figure: 3D graph showing a surface S, its unit normal vector n, and vector field F passing through the surface.

\(\vec{F} \cdot \vec{n}\) (the component perpendicularly thru surface)

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Calculating Surface Integrals of Vector Fields

How to calculate

\[ \iint_{S} \vec{F} \cdot d\vec{S} = \iint_{S} \vec{F} \cdot \vec{n} \, dS \]

?

Unit Normal Vector

Unit normal \( \vec{n} = \frac{\vec{r}_u \times \vec{r}_v}{|\vec{r}_u \times \vec{r}_v|} \) or \( \frac{\vec{r}_v \times \vec{r}_u}{|\vec{r}_v \times \vec{r}_u|} \) depending on which is in the positive direction.

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

Simplification

Then \( \iint_{S} \vec{F} \cdot \vec{n} \, dS \) becomes:

\[ \iint_{S} \vec{F} \cdot \frac{\vec{r}_u \times \vec{r}_v}{|\vec{r}_u \times \vec{r}_v|} |\vec{r}_u \times \vec{r}_v| \, dA = \iint_{R} \vec{F} \cdot (\vec{r}_u \times \vec{r}_v) \, dA \]

or \( \vec{r}_v \times \vec{r}_u \) depending on which one is positively oriented

Alternative (Explicit) Form

Alternative (explicit) form if \( z = z(x, y) \)

\( \vec{F} = \langle f, g, h \rangle \)

\[ \iint_{S} \vec{F} \cdot d\vec{S} = \iint_{R} (-f z_x - g z_y + h) \, dA \]

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Example: Surface Integral Calculation

Example: \( \vec{F} = \langle x, y, z \rangle \)

\( S \): plane \( 3x + 2y + z = 6 \) in the first octant.

The normal vector is considered positive upward.

Figure: 3D graph of the plane 3x+2y+z=6 in the first octant with normal vector arrows.

Figure: 2D projection of the surface onto the xy-plane, showing a triangular region R.

1. Parametrize Surface

Let \( u = x, v = y, z = 6 - 3x - 2y = 6 - 3u - 2v \)

Bounds:

\( 0 \le u \le 2 \)
\( 0 \le v \le 3 - \frac{3}{2}u \)

\[ \vec{r}(u, v) = \langle u, v, 6 - 3u - 2v \rangle \] \[ \vec{r}_u = \langle 1, 0, -3 \rangle \] \[ \vec{r}_v = \langle 0, 1, -2 \rangle \] \[ \vec{r}_u \times \vec{r}_v = \langle 3, 2, 1 \rangle \]

Stop. Is this pointing in the correct direction? Is this pointing upward? Yes, because z-component \( > 0 \).

So, use \( \langle 3, 2, 1 \rangle \) as correct normal.

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\[ \iint_{S} \vec{F} \cdot (\vec{r}_u \times \vec{r}_v) \, dA \]

Given the vector field and parametrization:

\[ \vec{F} = \langle x, y, z \rangle \quad \vec{r}(u,v) = \langle u, v, 6 - 3u - 2v \rangle \]

Substituting the components of \(\vec{r}(u,v)\) into \(\vec{F}\):

\(x = u\)
\(y = v\)
\(z = 6 - 3u - 2v\)

The integral becomes:

\[ = \int_{0}^{2} \int_{0}^{3 - \frac{3}{2}u} \underbrace{\langle u, v, 6 - 3u - 2v \rangle}_{\vec{F}} \cdot \underbrace{\langle 3, 2, 1 \rangle}_{\vec{r}_u \times \vec{r}_v} \, dv \, du \]

Evaluating the integral:

\[ = \dots = \boxed{18} \]

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Example

\(\vec{F} = \langle -y, x, 1 \rangle\)

\(S\): cylinder \(y = z^2\), \(0 \le x \le 3\), \(0 \le z \le 1\)

Normal is considered positive pointing toward positive \(y\)-axis

Figure: 3D graph showing axes x, y, z and a green surface y=z^2 with red normal vectors pointing in the positive y direction.

This is the one with component in pos. \(y\)-axis direction (we want this one)

Parametrize \(S\):

\(y = z^2\). We are given bounds in \(x\) and \(z\), so it's convenient to use them as parameters.

Let \(u = x\), \(v = z\), then \(y = z^2 = v^2\).

\[ \vec{r}(u,v) = \langle \underbrace{u}_{x}, \underbrace{v^2}_{y}, \underbrace{v}_{z} \rangle \]

With parameter bounds:

\(0 \le u \le 3\)
\(0 \le v \le 1\)

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\[ \vec{r}_u = \langle 1, 0, 0 \rangle \]\[ \vec{r}_v = \langle 0, 2v, 1 \rangle \]\[ \vec{r}_u \times \vec{r}_v = \langle 0, -1, 2v \rangle \]

is this normal pointing in the "positive" direction?

no, the y-component is negative so this points in negative y-axis direction

Fixing the Orientation

\[ \text{fix: } \vec{r}_v \times \vec{r}_u = \langle 0, 1, -2v \rangle \]

Calculating the surface integral:

\[ \iint_S \vec{F} \cdot (\vec{r}_v \times \vec{r}_u) \, dA = \int_0^3 \int_0^1 \langle -v^2, u, 1 \rangle \cdot \langle 0, 1, -2v \rangle \, dv \, du \]

\( = \dots = \)

\( \frac{3}{2} \)

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Example: Flux through a Sphere

\[ \vec{F} = \frac{-\langle x, y, z \rangle}{(x^2 + y^2 + z^2)^{3/2}} \]

S: sphere radius a
normal positive pointing outward

Parametrization of S

Using spherical coordinates:

\[ \rho = a, \quad u = \phi, \quad v = \theta \]\[ 0 \le u \le \pi, \quad 0 \le v \le 2\pi \]

\[ \vec{r}(u,v) = \langle a \sin u \cos v, \, a \sin u \sin v, \, a \cos u \rangle \]

Note: \( a \sin u \cos v \) corresponds to \( \rho \sin \phi \cos \theta \).

Partial Derivatives and Cross Product

\[ \vec{r}_u = \langle a \cos u \cos v, \, a \cos u \sin v, \, -a \sin u \rangle \]\[ \vec{r}_v = \langle -a \sin u \sin v, \, a \sin u \cos v, \, 0 \rangle \]\[ \vec{r}_u \times \vec{r}_v = \langle a^2 \sin^2 u \cos v, \, a^2 \sin^2 u \sin v, \, a^2 \cos u \sin u \rangle \]

"out" \( \rightarrow \) all components positive in first octant

here, all correct

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\[ \vec{F} = \frac{-\langle x, y, z \rangle}{(x^2 + y^2 + z^2)^{3/2}} = \frac{-\langle a \sin u \cos v, \ a \sin u \sin v, \ a \cos u \rangle}{a^3} \]

\[ \iint_S \vec{F} \cdot (\vec{r}_u \times \vec{r}_v) \, dA = \dots = \boxed{-4\pi} \]

see last page