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Flux and the Divergence Theorem

Let \(\mathbf{F} = 4x\mathbf{i} - z\mathbf{j} + x\mathbf{k}\). Compute Flux \[ \iint_S \mathbf{F} \cdot d\mathbf{S} \] where \(S\) is the union of the hemisphere \(x^2 + y^2 + z^2 = 1, z \ge 0\), and the base given by \(x^2 + y^2 \le 1, z = 0\). (Use the outward-pointing normal.)

Surface closed: can use Divergence Theorem

IF \(\vec{F}\) is defined everywhere inside the enclosed volume

Figure: 3D sketch of a hemisphere on the xy-plane with x, y, and z axes labeled.

Options:

A. \(\frac{2\pi}{3}\)
B. \(\frac{16\pi}{3}\)
C. \(\frac{8\pi}{3}\)
D. \(\frac{4\pi}{3}\)
E. \(\frac{-4\pi}{3}\)

Div. Theorem:

\[ \iint_S \vec{F} \cdot d\vec{S} = \iiint_E \text{div } \vec{F} \, dV \]

\(\text{div } \vec{F} = \nabla \cdot \langle 4x, -z, x \rangle = 4 + 0 + 0 = 4\)

\[ \iiint_E 4 \, dV = 4 \iiint_E dV = 4 \cdot \frac{1}{2} \cdot \frac{4}{3}\pi(1)^3 = \frac{16}{6}\pi = \frac{8\pi}{3} \]

volume of hemisphere

note if \(\text{div } \vec{F} = 1\) then \(\iint_S \vec{F} \cdot d\vec{S} = \text{volume}\) (Flux)

PAGE 2

very much like we can use Green's Theorem to find area

\[ \oint_C \vec{F} \cdot d\vec{r} = \iint_R (g_x - f_y) \, dA \]

Figure: 2D graph of a circular region R centered at the origin of the xy-axes.

if \(\vec{F}\) is such that \(g_x - f_y = 1\)

then \(\oint_C \vec{F} \cdot d\vec{r} = \text{area of } R\)

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Flux and Divergence Theorem

Find Flux \[ \iint_S \vec{F} \cdot \vec{n} \, dS \] for the region bounded between cylinder \( x^2 + z^2 = 4 \), \( (-2 \le y \le 2) \) and sphere \( x^2 + y^2 + z^2 = 1 \) given \( \vec{F} = \langle 3x + y, 2y + z, z \rangle \) including \( y = 2 \) and \( y = -2 \).

A. 0
B. \( \frac{44\pi}{3} \)
C. \( 88\pi \)
D. \( \frac{188\pi}{3} \)
E. \( 376\pi \)

Figure: 3D graph with x, y, z axes showing a cylinder along the y-axis from -2 to 2 and a sphere at the origin.

flux through the surface bounding the space between sphere and capped cylinder

Divergence Theorem for hollowed volume

\[ \iint_S \vec{F} \cdot \vec{n} \, dS = \iiint_{\text{outside}} \text{div} \vec{F} \, dV - \iiint_{\text{inside}} \text{div} \vec{F} \, dV \]

\[ \text{div} \vec{F} = 3 + 2 + 1 = 6 \]

\[ \iiint_{\text{cyl}} 6 \, dV - \iiint_{\text{sph}} 6 \, dV = 6 \left( \iiint_{\text{cyl}} dV - \iiint_{\text{sph}} dV \right) \]

vol. of cylinder

\[ = 6 \left( \pi(2)^2 \cdot 4 - \frac{4}{3}\pi \right) = 6\pi \left( 16 - \frac{4}{3} \right) = 6\pi \left( \frac{44}{3} \right) = 88\pi \]

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Particle Position and Arc Length

The position of a particle is given by \( \vec{r}(t) = \langle 2t, 1 - 2t, 5 + t \rangle \), starting when \( t = 0 \). After the particle has gone a distance of 3, the \( x \)-coordinate is

A. \( \frac{1}{3} \)
B. 3
C. \( \frac{1}{2} \)
D. 2
E. 1

Figure: A diagram of a vector path r(t) starting at t=0, showing an arc length s=3 reaching a point (x, y, z) at t=?.

Solution

length: \[ s(t) = \int_0^t |\vec{r}'| \, du \]

\[ = \int_0^t 3 \, du = 3t \]

\[ \vec{r}' = \langle 2, -2, 1 \rangle \]

\[ |\vec{r}'| = \sqrt{4 + 4 + 1} = 3 \]

want \( s = 3 \rightarrow 3 = 3t \rightarrow t = 1 \)

at \( t = 1 \), \( \vec{r}(1) = \langle 2, -1, 6 \rangle \)

The x-coordinate is 2.

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Directional Derivatives and Gradients

19 from practice problems set

Problem Statement

Find directional derivative of \( f(x,y) = 5 - 4x^2 - 3y \) at \( (x,y) \) towards the origin.

\[ D_{\vec{u}}f = \vec{\nabla}f \cdot \vec{u} \]

Figure: Coordinate axes with a vector starting at point (x,y) and pointing to the origin, labeled <-x, -y>.

Note: \( \vec{\nabla}f \) is the gradient and \( \vec{u} \) is the unit vector specifying direction.

Solution

The unit vector \( \vec{u} \) towards the origin is:

\[ \vec{u} = \frac{\langle -x, -y \rangle}{\sqrt{x^2 + y^2}} \]

The gradient of \( f \) is:

\[ \vec{\nabla}f = \langle -8x, -3 \rangle \]

Calculating the directional derivative:

\[ D_{\vec{u}}f = \langle -8x, -3 \rangle \cdot \frac{\langle -x, -y \rangle}{\sqrt{x^2 + y^2}} = \frac{8x^2 + 3y}{\sqrt{x^2 + y^2}} \]

Properties of the Gradient \( \vec{\nabla}f \)

Direction? Greatest increase \( \rightarrow \perp \) to level curve.
Magnitude? Maximum directional derivative.
No \( D_{\vec{u}}f \) if going \( \perp \vec{\nabla}f \) or parallel to level curve.

PAGE 6

Extrema of Functions of Two Variables

Problem #24

The function \( f(x,y) = 2x^3 - 6xy - 3y^2 \). Find max/min/saddle pt.

1. Find Critical Points

Set partial derivatives to zero: \( f_x = 0 \) AND \( f_y = 0 \).

\[ f_x = 6x^2 - 6y = 0 \implies x^2 = y \implies x = y, -y \]\[ f_y = -6x - 6y = 0 \implies x = -y \]

Substitute \( x = -y \) into \( f_x = 0 \):

\[ 6(-y)^2 - 6y = 0 \]\[ 6y^2 - 6y = 0 \]\[ 6y(y - 1) = 0 \implies y = 0, y = 1 \]

Corresponding x-values: \( x = 0, x = -1 \)

Critical Points (cp): \( (0,0), (-1, 1) \)

2. Second Derivative Test

The discriminant \( D \) is defined as:

\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]

\[ f_{xx} = 12x \]\[ f_{yy} = -6 \]

\[ f_{xy} = -6 \]

\[ D(x,y) = (12x)(-6) - (-6)^2 = -72x - 36 \]

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Second Derivative Test Analysis

At the point \((0,0)\):

\[D = -36 < 0 \rightarrow \text{saddle pt}\]

At the point \((-1,1)\):

\[D = 72 - 36 > 0 \rightarrow \text{max or min}\]

Check \(f_{xx}\):

\[f_{xx}(-1,1) < 0 \rightarrow \text{max}\]

If \(D = 0 \rightarrow \text{inconclusive, this test fails}\)

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Problem 14: Mixed Partial Derivatives

Given the function:

\[f(x,y) = \cos(xy)\]

Find the mixed partial derivative:

\[\frac{\partial^2 f}{\partial x \partial y} = ?\]

Note: Differentiate with respect to \(y\) first, then \(x\). When differentiating with respect to \(y\), treat \(x\) as a "constant".

Step 1: First Partial Derivative with respect to \(y\)

\[\frac{\partial}{\partial y} (\cos(xy)) = -\sin(xy) \cdot x = -x \sin(xy)\]

Step 2: Second Partial Derivative with respect to \(x\)

Apply the product rule to \(\frac{\partial}{\partial x} [(-x) \cdot \sin(xy)]\), treating \(y\) as a "constant":

\[= (-x) \cdot \frac{\partial}{\partial x} \sin(xy) + \sin(xy) \cdot \frac{\partial}{\partial x} (-x)\]

\[= (-x) \cdot \cos(xy) \cdot y + \sin(xy) \cdot (-1)\]

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Vector Fields and Potential Functions

A potential for the vector field \(\mathbf{F}(x, y) = (3x^2y + 2xy, \, x^3 + x^2)\) is

A. \(xy^3 + xy^2 + C\)
B. \(x^3y + x^2y + C\)
C. \(x^2y^3 + xy^3 + C\)
D. \(x^2y^3 + x^3 + C\)
E. A potential for \(\mathbf{F}\) does not exist.

Solution Derivation

\[ \nabla \phi = \vec{F} \]\[ \langle \phi_x, \phi_y \rangle = \langle 3x^2y + 2xy, \, x^3 + x^2 \rangle \]

From the components, we have:

\(\phi_x = 3x^2y + 2xy\)
\(\phi_y = x^3 + x^2\)

To find \(\phi\), we integrate \(\phi_x\) with respect to \(x\):

\[ \phi = \int (3x^2y + 2xy) dx = x^3y + x^2y + h(y) \]

Note: This expression represents what we know about \(\phi\) so far.

Now, take the partial of our \(\phi\) with respect to \(y\):

\[ \phi_y = x^3 + x^2 + \frac{dh}{dy} \]\[ \text{It must be equal to } \phi_y = x^3 + x^2 \text{ (from } \vec{F} \text{)} \]

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Comparing the two expressions for \(\phi_y\):

\[ \text{so, } \frac{dh}{dy} = 0 \quad \text{which means } h = C \text{ (numerical constant)} \]

Substituting back into our expression for \(\phi\):

\[ \phi = x^3y + x^2y + C \]

Therefore, the correct option is B.

Fundamental Theorem of Line Integrals

\(\phi\) is important in the Fundamental Theorem of Line Integrals:

\[ \int_C \nabla \phi \cdot d\vec{r} = \phi(\text{end}) - \phi(\text{start}) \quad : \text{path independent} \]\[ \int_C \vec{F} \cdot d\vec{r} \text{ if } \vec{F} \text{ is conservative} \]

\[ \vec{F} \text{ is conservative } \implies \text{curl } \vec{F} = \vec{0} \]