Final Review (05/01)

Accessible transcription generated on 5/1/2026

Original Notes

Page 1

Final Review (05/01)

Compute Line Integrals

FTCLI (Fundamental Theorem of Calculus for Line Integrals) \[ \int_C \nabla \phi \cdot d\vec{r} = \phi(\text{end}) - \phi(\text{start}) \] if is given a conservative \( \vec{F} \)
Stokes' Theorem \[ \oint_C \vec{F} \cdot d\vec{r} = \iint_S \text{curl } \vec{F} \cdot \vec{n} \, dS \]
A 3D diagram representing Stokes' Theorem. It shows an open, wavy surface S in space. The surface is bounded by a closed curve C at its base. Small vertical line segments are drawn on the surface S to indicate its orientation and normal vectors.
Visual Description: A 3D diagram representing Stokes' Theorem. It shows an open, wavy surface S in space. The surface is bounded by a closed curve C at its base. Small vertical line segments are drawn on the surface S to indicate its orientation and normal vectors.
Green's: Special case

Flat in 2D.

A 2D diagram representing Green's Theorem as a special case of Stokes' Theorem. It depicts a flat, irregular region S in a plane, shaded with diagonal hatch lines. The region is enclosed by a boundary curve C.
Visual Description: A 2D diagram representing Green's Theorem as a special case of Stokes' Theorem. It depicts a flat, irregular region S in a plane, shaded with diagonal hatch lines. The region is enclosed by a boundary curve C.

Divergence; Compute Surface Integral

Divergence Theorem
A diagram representing the Divergence Theorem. It shows a closed, 3D blob-like volume D bounded by a surface S. Multiple arrows, labeled with the unit normal vector n, point outward from various locations on the surface S, representing the outward flux from the volume.
Visual Description: A diagram representing the Divergence Theorem. It shows a closed, 3D blob-like volume D bounded by a surface S. Multiple arrows, labeled with the unit normal vector n, point outward from various locations on the surface S, representing the outward flux from the volume.
\[ \iint_S \vec{F} \cdot \vec{n} \, dS = \iiint_D \text{div } \vec{F} \, dV \]

Page 2

To compute line integrals \(\int_C \vec{F} \cdot d\vec{r}\)

Is \(\vec{F}\) conservative?

  • Yes: Use FTCLI (Fundamental Theorem for Line Integrals).
  • NO: Is \(C\) a closed curve?
    • Yes in 2D: Use Greens theorem.
    • Yes in 3D: Use Stokes theorem.
    • NO: Default: parametrize \(C\) & solve.

Page 3

To compute Surface integral \[\iint_S \vec{F} \cdot \vec{n} \, dS\]

Determine the appropriate method for calculation based on the following logic flow:

Is \(\vec{F}\) a Curl Field? (i.e., is \(\vec{F} = \text{curl}(\vec{G}) = \nabla \times \vec{G}\)?)

  • Yes: Use Stokes' Theorem.
  • No: Proceed to the next check:
    • Is \(S\) a closed surface?
      • Yes: Use Divergence Theorem.
      • No: Use the Default: parametrization.

Page 4

Applying Divergence Theorem to Open Surfaces

A three-dimensional technical sketch illustrating a method for calculating flux through a non-closed surface. It shows an irregular, open upper surface S, which is noted as being 'hard to parametrize'. To create a closed volume D, a flat horizontal base surface S_1 is added at the bottom. An upward-pointing arrow on S and a downward-pointing arrow on S_1 indicate the outward-pointing normal vectors for the closed region. Parallel hatching lines are used on the surfaces to provide a 3D perspective of the enclosed volume.
Visual Description: A three-dimensional technical sketch illustrating a method for calculating flux through a non-closed surface. It shows an irregular, open upper surface S, which is noted as being 'hard to parametrize'. To create a closed volume D, a flat horizontal base surface S_1 is added at the bottom. An upward-pointing arrow on S and a downward-pointing arrow on S_1 indicate the outward-pointing normal vectors for the closed region. Parallel hatching lines are used on the surfaces to provide a 3D perspective of the enclosed volume.

When a surface \( S \) has the following characteristics:

  • \( S \): Something hard to parametrize
  • Not closed
  • \( \vec{F} \): Not a curl field

We can apply the Divergence Theorem by closing the surface \( S \) with a simpler surface (such as the flat base \( S_1 \) shown in the diagram) to enclose a volume \( D \). The theorem relates the triple integral over the volume to the total flux through the boundary surfaces:

\[ \iiint_D \text{div } \vec{F} \, dV = \iint_S \vec{F} \cdot d\vec{S} + \iint_{S_1} \vec{F} \cdot d\vec{S} \]

In the notes, the term \( \iint_S \vec{F} \cdot d\vec{S} \) is circled in red with a question mark (\(?\)) beneath it. This indicates that the goal of this setup is typically to find the flux through the complex surface \( S \) by instead calculating the volume integral and the flux through the simpler surface \( S_1 \).

Page 5

Example 1

Let \( S \) be the surface \( z = \sqrt{4 - x^2 - y^2} \) inside the cylinder \( x^2 + y^2 = 1 \).
Let \( \vec{F} = \langle yz, -xz, xy \rangle \).
Compute \( \iint_S (\vec{\nabla} \times \vec{F}) \cdot d\vec{S} \).

A 3D coordinate system diagram illustrating a Stokes' theorem problem. It depicts a hemisphere defined by \( z = \sqrt{4 - x^2 - y^2} \), representing part of a sphere with radius 2. A vertical cylinder with radius 1, defined by \( x^2 + y^2 = 1 \), passes through the sphere. The surface S is identified as the portion of the hemisphere contained within this cylinder and is highlighted with blue diagonal shading. The boundary curve C is the circle formed at the intersection of the cylinder and the hemisphere. The x, y, and z axes are clearly shown, with the cylinder extending vertically along the z-axis.
Visual Description: A 3D coordinate system diagram illustrating a Stokes' theorem problem. It depicts a hemisphere defined by \( z = \sqrt{4 - x^2 - y^2} \), representing part of a sphere with radius 2. A vertical cylinder with radius 1, defined by \( x^2 + y^2 = 1 \), passes through the sphere. The surface S is identified as the portion of the hemisphere contained within this cylinder and is highlighted with blue diagonal shading. The boundary curve C is the circle formed at the intersection of the cylinder and the hemisphere. The x, y, and z axes are clearly shown, with the cylinder extending vertically along the z-axis.

Strategy: Use Stokes' Theorem!

By Stokes' Theorem: \[ \iint_S (\vec{\nabla} \times \vec{F}) \cdot d\vec{S} = \oint_C \vec{F} \cdot d\vec{r} \]

1. Parametrize \( C \):
The boundary curve \( C \) is the intersection of \( x^2 + y^2 = 1 \) and \( z = \sqrt{4 - x^2 - y^2} \).
Substituting the cylinder equation into the surface equation: \[ z = \sqrt{4 - 1} = \sqrt{3} \] The parametrization is: \[ \vec{r}(t) = \langle \cos t, \sin t, \sqrt{3} \rangle, \quad 0 \le t \le 2\pi \]

2. Compute the line integral: \[ \oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt \] \[ = \int_0^{2\pi} \langle \sqrt{3} \sin t, -\sqrt{3} \cos t, \cos t \sin t \rangle \cdot \langle -\sin t, \cos t, 0 \rangle \, dt \] \[ = \dots \]

Page 6

Example 2

\( C \) is the boundary of \( z = 6 - 3x - y \) in the \( 1^{\text{st}} \) octant.
Given the vector field \( \vec{F} = \langle y + 2z, -x, -2x \rangle \), compute \( \oint_C \vec{F} \cdot d\vec{r} \).

Since \( C \) is a closed curve in 3D, we can Apply Stokes theorem: \[ \oint_C \vec{F} \cdot d\vec{r} = \iint_S \text{curl } \vec{F} \cdot d\vec{S} \]

A 3D coordinate system graph showing the surface S in the first octant. The surface S is a triangular part of a plane with vertices on the x, y, and z axes. The boundary curve C follows the edges of this triangle, with arrows indicating a counter-clockwise orientation when viewed from above. The surface area of the triangle is shaded with horizontal lines.
Visual Description: A 3D coordinate system graph showing the surface S in the first octant. The surface S is a triangular part of a plane with vertices on the x, y, and z axes. The boundary curve C follows the edges of this triangle, with arrows indicating a counter-clockwise orientation when viewed from above. The surface area of the triangle is shaded with horizontal lines.

First, compute the curl of \( \vec{F} \): \[ \text{curl } \vec{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial/\partial x & \partial/\partial y & \partial/\partial z \\ y+2z & -x & -2x \end{vmatrix} = \langle 0, +4, -2 \rangle \]

The surface \( S \) is defined by \( z = 6 - 3x - y \). Treating this as a plane equation \( 3x + y + z = 6 \), the normal vector is: \[ \vec{n} = \langle 3, 1, 1 \rangle \]

Alternatively, using a parametrization: \[ \vec{r}(u, v) = \langle u, v, 6 - 3u - v \rangle \] \[ \vec{r}_u = \langle 1, 0, -3 \rangle \] \[ \vec{r}_v = \langle 0, 1, -1 \rangle \] \[ \vec{r}_u \times \vec{r}_v = \langle 3, 1, 1 \rangle \]

The region of integration \( D \) is the projection of the surface \( S \) onto the \( xy \)-plane.

A 2D coordinate graph showing the triangular region D in the uv-plane (or xy-plane). The region is bounded by the vertical axis (intercept at 6), the horizontal axis (labeled 'u' with an intercept at 2), and the line equation 3x + y = 6. The triangle has vertices at (0,0), (2,0), and (0,6). The interior of the triangle is shaded with diagonal lines.
Visual Description: A 2D coordinate graph showing the triangular region D in the uv-plane (or xy-plane). The region is bounded by the vertical axis (intercept at 6), the horizontal axis (labeled 'u' with an intercept at 2), and the line equation 3x + y = 6. The triangle has vertices at (0,0), (2,0), and (0,6). The interior of the triangle is shaded with diagonal lines.

Evaluate the surface integral over the region \( D \): \[ = \iint_D \langle 0, +4, -2 \rangle \cdot \langle 3, 1, 1 \rangle dA \] \[ = \iint_D (0 + 4 - 2) dA = 2 \iint_D 1 dA \] \[ = 2 \times \text{Area of } D \] \[ = 2 \times \frac{1}{2}(2 \times 6) \] \[ = 12 \]

Page 7

Example 3

\( \vec{F} = \langle x^3, y^3, z^3 \rangle \), \( S \) is the closed surface \( z = \sqrt{4 - x^2 - y^2} \) together with \( x^2 + y^2 \le 4 \) on \( z = 0 \).

Evaluate \( \iint_S \vec{F} \cdot d\vec{S} \).

A 3D coordinate system diagram showing a solid upper hemisphere representing the region D. The vertical axis is labeled 'z' and the horizontal axes represent the x-y plane. The hemisphere is centered at the origin, with its base being a circle in the xy-plane defined by the equation \(x^2 + y^2 \le 4\). The upper curved boundary is defined by \(z = \sqrt{4 - x^2 - y^2}\). The interior of the hemisphere is shaded with diagonal hatch marks to indicate a volume, and it is labeled with a 'D'. A red line segment is drawn from the origin to the surface of the hemisphere, indicating the radius.
Visual Description: A 3D coordinate system diagram showing a solid upper hemisphere representing the region D. The vertical axis is labeled 'z' and the horizontal axes represent the x-y plane. The hemisphere is centered at the origin, with its base being a circle in the xy-plane defined by the equation \(x^2 + y^2 \le 4\). The upper curved boundary is defined by \(z = \sqrt{4 - x^2 - y^2}\). The interior of the hemisphere is shaded with diagonal hatch marks to indicate a volume, and it is labeled with a 'D'. A red line segment is drawn from the origin to the surface of the hemisphere, indicating the radius.

\( S \) is closed \( \implies \) Apply Divergence Theorem.

\[ \text{div} \vec{F} = \nabla \cdot \vec{F} = f_x + g_y + h_z \] \[ = 3x^2 + 3y^2 + 3z^2 \]

By the Divergence Theorem: \[ \iint_S \vec{F} \cdot d\vec{S} = \iiint_D \text{div} \vec{F} \, dV = \iiint_D 3(x^2 + y^2 + z^2) \, dV \]

Use spherical coordinates: \[ = \int_0^{2\pi} \int_0^{\pi/2} \int_0^2 3\rho^2 \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]

Page 8

Example 4

\( x = y^2 + z^2, \quad x \leq 3 \)
\( \vec{n} \) towards positive \( x \)-axis
\( \vec{F} = \langle 4x, y, z \rangle \)

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

A 3D coordinate system diagram showing a paraboloid surface. The x-axis points diagonally towards the lower-left, the y-axis is horizontal to the right, and the z-axis is vertical. The paraboloid is defined by the equation x = y^2 + z^2, with its vertex at the origin (0,0,0) and opening along the positive x-axis. The surface is truncated at the plane x = 3, where it ends in a circular cross-section. The area of the paraboloid surface between x = 0 and x = 3 is shaded with blue diagonal lines, and a label 'x=3' points to the boundary circle at the end.
Visual Description: A 3D coordinate system diagram showing a paraboloid surface. The x-axis points diagonally towards the lower-left, the y-axis is horizontal to the right, and the z-axis is vertical. The paraboloid is defined by the equation x = y^2 + z^2, with its vertex at the origin (0,0,0) and opening along the positive x-axis. The surface is truncated at the plane x = 3, where it ends in a circular cross-section. The area of the paraboloid surface between x = 0 and x = 3 is shaded with blue diagonal lines, and a label 'x=3' points to the boundary circle at the end.

Parametrization: \[ \vec{r}(u, v) = \langle u^2 + v^2, u, v \rangle \] \[ \vec{r}_u = \langle 2u, 1, 0 \rangle \] \[ \vec{r}_v = \langle 2v, 0, 1 \rangle \] \[ \vec{r}_u \times \vec{r}_v = \langle 1, -2u, -2v \rangle \]

Vector field in terms of parameters: \[ \vec{F}(\vec{r}(u, v)) = \langle 4u^2 + 4v^2, u, v \rangle \]

Bounds for \( u, v \): Since \( x \leq 3 \), we have \( u^2 + v^2 \leq 3 \).

Integral Setup: \[ \iint_{u^2 + v^2 \leq 3} \langle 4u^2 + 4v^2, u, v \rangle \cdot \langle 1, -2u, -2v \rangle \, dA = \]

Page 9

Outward Flux

Outward flux: \( \vec{n} \) is pointing outwards.

\( S \) is a closed surface.

A hand-drawn technical sketch of a hemispherical closed surface S. The diagram depicts a three-dimensional dome shape resting on a flat, circular base. The interior of the solid is shaded with light, parallel diagonal lines. Three arrows representing outward-pointing unit normal vectors (n) are shown: one arrow points straight up from the top of the dome, one arrow points downward and to the left from the edge of the circular base, and another arrow points downward and to the right from the opposite edge of the base.
Visual Description: A hand-drawn technical sketch of a hemispherical closed surface S. The diagram depicts a three-dimensional dome shape resting on a flat, circular base. The interior of the solid is shaded with light, parallel diagonal lines. Three arrows representing outward-pointing unit normal vectors (n) are shown: one arrow points straight up from the top of the dome, one arrow points downward and to the left from the edge of the circular base, and another arrow points downward and to the right from the opposite edge of the base.
A hand-drawn diagram of a 3D cube representing a closed surface S. The cube is sketched using perspective, with its faces textured with diagonal hatching lines. Multiple arrows are drawn originating from the center of various faces to represent the outward-pointing unit normal vector (n). An arrow points straight up from the top face, one points directly to the right from the right face, one points directly to the left from the left face, and a fourth arrow points straight down from the bottom face, illustrating the convention that normal vectors for a closed volume always point away from the interior.
Visual Description: A hand-drawn diagram of a 3D cube representing a closed surface S. The cube is sketched using perspective, with its faces textured with diagonal hatching lines. Multiple arrows are drawn originating from the center of various faces to represent the outward-pointing unit normal vector (n). An arrow points straight up from the top face, one points directly to the right from the right face, one points directly to the left from the left face, and a fourth arrow points straight down from the bottom face, illustrating the convention that normal vectors for a closed volume always point away from the interior.

Page 10

Example 5

Example: Calculating Work Along a Path

Given the vector field:

\[ \vec{F} = \left\langle -\frac{x}{2}, -\frac{y}{2}, \frac{1}{4} \right\rangle \]

And the parameterized curve:

\[ \vec{r}(t) = \langle \cos t, \sin t, t \rangle \]

Calculate the Work done from the point \((1, 0, 0)\) to \((-1, 0, 3\pi)\).

First, determine the integration limits for the parameter \( t \):

  • At point \((1, 0, 0)\), we find \( t = 0 \).
  • At point \((-1, 0, 3\pi)\), we find \( t = 3\pi \).

The formula for work is the line integral of the vector field along the curve \( C \):

\[ \text{Work} = \int_{C} \vec{F} \cdot d\vec{r} = \int_{0}^{3\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt \]

Substituting the parameterized components into the vector field and finding the derivative of the path:

  • \(\vec{F}(\vec{r}(t)) = \left\langle -\frac{\cos t}{2}, -\frac{\sin t}{2}, \frac{1}{4} \right\rangle\)
  • \(\vec{r}'(t) = \langle -\sin t, \cos t, 1 \rangle\)

Substituting these into the integral yields:

\[ = \int_{0}^{3\pi} \left\langle -\frac{\cos t}{2}, -\frac{\sin t}{2}, \frac{1}{4} \right\rangle \cdot \langle -\sin t, \cos t, 1 \rangle \, dt \] \[ = \dots \]

Page 11

Example 6

\[ \phi = x e^{xyz} \]

A hand-drawn 3D wireframe sketch of a cube. The label 'D =' is written to the left of the cube, indicating it represents the solid region D. The sketch uses two overlapping squares connected at their corners with diagonal lines to represent the volume in perspective.
Visual Description: A hand-drawn 3D wireframe sketch of a cube. The label 'D =' is written to the left of the cube, indicating it represents the solid region D. The sketch uses two overlapping squares connected at their corners with diagonal lines to represent the volume in perspective.

Evaluate the following surface integral: \[ \iint_{S = \partial D} (\vec{\nabla} \times \vec{\nabla} \phi) \cdot \vec{n} \, dS \]

There are two primary methods shown in the notes to evaluate this integral:

1. Using Stokes' Theorem: Applying Stokes' Theorem relates the surface integral of the curl to a line integral around the boundary curve \(C\). \[ \iint_{S = \partial D} (\vec{\nabla} \times \vec{\nabla} \phi) \cdot \vec{n} \, dS \xrightarrow{\text{Stokes}} \oint_{C = \partial S} \vec{\nabla} \phi \cdot d\vec{r} = 0 \] Since \(S = \partial D\) is the closed boundary surface of a solid region \(D\), it has no boundary curve (\(C = \emptyset\)), thus the integral is zero.

2. Using the Divergence Theorem: Alternatively, we can apply the divergence theorem. We use the fundamental vector identity that the divergence of a curl is always zero: \[ \text{div}(\text{curl } \vec{F}) = 0 \] Where in this case \(\vec{F} = \vec{\nabla} \phi\). Converting the surface integral to a volume integral over region \(D\): \[ \iiint_{D} 0 \, dV \] \[ \mathbf{= 0} \]