Exam 2: Review

Accessible transcription generated on 4/20/2026

Original Notes

Page 1

Exam 2: Review

  • Exam 2: Tomorrow (Tuesday 04/21), 8 PM
  • Office Hours:
    • Today (04/20): 10:15 AM – 11:45 AM (MATH 842)
    • 12:15 PM – 1:15 PM (Windsor)
    • Tomorrow (04/21): 12:00 PM – 1:15 PM (MATH 842)
  • No Class on Friday (04/24)

Page 2

Constrained Optimization

Find minimum value of \( 2x + 3y + 2 \) given \( 2x^2 + 5xy + 4y^2 = 28 \).

Objective: \( f(x, y) = 2x + 3y + 2 \)
Constraint: \( g(x, y) = 2x^2 + 5xy + 4y^2 = 28 \)

Lagrange Multipliers: \( \vec{\nabla} f = \lambda \vec{\nabla} g \rightsquigarrow \langle f_x, f_y \rangle = \lambda \langle g_x, g_y \rangle \)

  1. \( f_x = \lambda g_x \rightsquigarrow 2 = \lambda (4x + 5y) \)
  2. \( f_y = \lambda g_y \rightsquigarrow 3 = \lambda (5x + 8y) \)
  3. \( g(x, y) = 28 \)

Divide equation (1) by equation (2):
\[ \frac{2}{3} = \frac{4x + 5y}{5x + 8y} \] \[ 10x + 16y = 12x + 15y \] \[ y = 2x \]

Plug into equation (3):
\[ 2x^2 + 5x(2x) + 4(2x)^2 = 28 \] \[ 2x^2 + 10x^2 + 16x^2 = 28 \] \[ 28x^2 = 28 \] \[ x = \pm 1 \]

Points: \( (1, 2) \) and \( (-1, -2) \)

Evaluate the objective function at these points:

  • \( f(1, 2) = 2(1) + 3(2) + 2 = 2 + 6 + 2 = 10 \)
  • \( f(-1, -2) = 2(-1) + 3(-2) + 2 = -2 - 6 + 2 = -6 \)

The minimum value is -6.

Page 3

Change of Order of Integration Example

Evaluate \[ \int_{0}^{1} \int_{\sqrt{x}}^{1} e^{y^3} \, dy \, dx \]
\( e^{y^3} \) is not easy to integrate with respect to \( y \). Switch the order of integration.
Given bounds: \( \sqrt{x} \le y \le 1, 0 \le x \le 1 \)
A Cartesian coordinate graph showing the region of integration. The x-axis is labeled from 0 to 1. A blue curve is plotted, defined by the equation y = sqrt(x), which is also labeled as x = y squared. A horizontal blue line is drawn at y = 1. The region bounded between the y-axis (x=0), the horizontal line y=1, and the curve x = y squared is shaded with diagonal lines. A purple horizontal line segment with markers at both ends is drawn across the shaded region at a representative height 'y', indicating a horizontal slice starting at the y-axis (x=0) and ending at the curve (x=y squared). This visualizes the new integration limits for x.
Visual Description: A Cartesian coordinate graph showing the region of integration. The x-axis is labeled from 0 to 1. A blue curve is plotted, defined by the equation y = sqrt(x), which is also labeled as x = y squared. A horizontal blue line is drawn at y = 1. The region bounded between the y-axis (x=0), the horizontal line y=1, and the curve x = y squared is shaded with diagonal lines. A purple horizontal line segment with markers at both ends is drawn across the shaded region at a representative height 'y', indicating a horizontal slice starting at the y-axis (x=0) and ending at the curve (x=y squared). This visualizes the new integration limits for x.
From the region shown above, the new bounds are: \[ 0 \le x \le y^2, 0 \le y \le 1 \]
The integral with the changed order of integration is: \[ \int_{0}^{1} \int_{0}^{y^2} e^{y^3} \, dx \, dy \] Evaluating the inner integral with respect to \( x \): \[ = \int_{0}^{1} \left[ x e^{y^3} \right]_{0}^{y^2} \, dy \] \[ = \int_{0}^{1} y^2 e^{y^3} \, dy \] To solve the remaining integral, use substitution. Let \( u = y^3 \), then \( du = 3y^2 \, dy \), which implies \( y^2 \, dy = \frac{1}{3} du \). The limits of integration for \( u \) remain \( 0 \) to \( 1 \). \[ = \frac{1}{3} \int_{0}^{1} e^u \, du \] \[ = \frac{1}{3} [e^u]_{0}^{1} \] \[ = \frac{1}{3} [e - 1] \]

Page 4

Rewrite in polar

\[ \int_{0}^{1} \int_{-\sqrt{1-y^2}}^{0} (x^2+y^2)^{1/2} dx dy \]
Coordinate plane diagram showing a region of integration in the second quadrant. The x and y axes are clearly labeled. A quarter-circle region with a radius of 1 is shaded with diagonal blue lines, extending from the positive y-axis to the negative x-axis. A red radius line starts from the origin (0,0) and extends to a point on the circular arc. A red curved arrow indicates the angle theta, starting from the positive x-axis and rotating counter-clockwise to the red radius line, showing that the angular range for this region is from pi/2 to pi.
Visual Description: Coordinate plane diagram showing a region of integration in the second quadrant. The x and y axes are clearly labeled. A quarter-circle region with a radius of 1 is shaded with diagonal blue lines, extending from the positive y-axis to the negative x-axis. A red radius line starts from the origin (0,0) and extends to a point on the circular arc. A red curved arrow indicates the angle theta, starting from the positive x-axis and rotating counter-clockwise to the red radius line, showing that the angular range for this region is from pi/2 to pi.

Based on the limits of the original double integral, we identify the region in Cartesian coordinates as:

\[ -\sqrt{1-y^2} \leq x \leq 0, \quad 0 \leq y \leq 1 \]

To determine the region's geometry for the polar conversion, we analyze the lower boundary of \( x \):

\[ -\sqrt{1-y^2} = x \] \[ x^2 + y^2 = 1 \]

Using the visual representation from the diagram, we establish the limits of integration in polar coordinates:

\[ 0 \leq r \leq 1, \quad \frac{\pi}{2} \leq \theta \leq \pi \]

The standard substitutions for polar coordinates are applied:

\[ dx dy = r dr d\theta \] \[ x^2 + y^2 = r^2 \]

Substituting these values into the integral and evaluating the result:

\[ \int_{\frac{\pi}{2}}^{\pi} \int_{0}^{1} (r^2)^{1/2} \cdot r \, dr d\theta = \frac{1}{3} \cdot \frac{\pi}{2} = \frac{\pi}{6} \]

Page 5

Triple Integrals: Volume in Spherical and Cylindrical Coordinates

Find the volume of the region inside \( x^2 + y^2 + z^2 = 4 \) (Sphere of radius 2) and \( z = \sqrt{3x^2 + 3y^2} \) (Cone).
A 3D coordinate system diagram showing a sphere and a cone. The sphere, defined by \( x^2 + y^2 + z^2 = 4 \), has a radius of 2 and is centered at the origin. An upward-opening cone, defined by \( z = \sqrt{3x^2 + 3y^2} \), intersects the sphere. The volume shared by the interior of the cone and the sphere is shaded with horizontal lines. A red vertical line segment with points at the boundary indicates the z-bounds for a point within the region, extending from the cone surface up to the spherical cap. Dashed lines project the circular intersection of the cone and sphere onto the xy-plane.
Visual Description: A 3D coordinate system diagram showing a sphere and a cone. The sphere, defined by \( x^2 + y^2 + z^2 = 4 \), has a radius of 2 and is centered at the origin. An upward-opening cone, defined by \( z = \sqrt{3x^2 + 3y^2} \), intersects the sphere. The volume shared by the interior of the cone and the sphere is shaded with horizontal lines. A red vertical line segment with points at the boundary indicates the z-bounds for a point within the region, extending from the cone surface up to the spherical cap. Dashed lines project the circular intersection of the cone and sphere onto the xy-plane.
To find the volume using Cartesian or Cylindrical coordinates, we first determine the bounds for \( z \). From the equations given: \[ \sqrt{3x^2 + 3y^2} \le z \le \sqrt{4 - x^2 - y^2} \] The xy bounds are determined by the "shadow" of the region, which is a disk with a boundary formed by the intersection of the sphere and the cone. We find this intersection by setting the \( z \)-expressions equal: \[ \sqrt{3x^2 + 3y^2} = \sqrt{4 - x^2 - y^2} \] \[ 3x^2 + 3y^2 = 4 - x^2 - y^2 \] \[ 4x^2 + 4y^2 = 4 \] \[ x^2 + y^2 = 1 \]
A 2D coordinate plane diagram showing the projection of the intersection region onto the xy-plane. It depicts a shaded unit circle centered at the origin, labeled with the equation \( x^2 + y^2 = 1 \). The x and y axes are clearly marked, and the circle intersects the axes at 1 and -1.
Visual Description: A 2D coordinate plane diagram showing the projection of the intersection region onto the xy-plane. It depicts a shaded unit circle centered at the origin, labeled with the equation \( x^2 + y^2 = 1 \). The x and y axes are clearly marked, and the circle intersects the axes at 1 and -1.
Converting to polar coordinates for the \( xy \)-plane: \[ dx \, dy = r \, dr \, d\theta \] The limits for the disk \( x^2 + y^2 \le 1 \) are: \[ 0 \le r \le 1 \] \[ 0 \le \theta \le 2\pi \]

Page 6

Cartesian!

\[ \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{\sqrt{3x^2+3y^2}}^{\sqrt{4-x^2-y^2}} 1 \, dz \, dy \, dx \]

Cylindrical!

\[ \int_{0}^{2\pi} \int_{0}^{1} \int_{\sqrt{3}r}^{\sqrt{4-r^2}} 1 \cdot r \, dz \, dr \, d\theta \]

Spherical!

\( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \)
Bounds:
  • \( 0 \leq \rho \leq 2 \)
  • \( 0 \leq \phi \leq \) hit cone
  • \( 0 \leq \theta \leq 2\pi \)
Derivation of the cone boundary angle \(\phi\):

Given the Cartesian equation for the cone boundary:

\[ z = \sqrt{3(x^2+y^2)} \]

Using spherical coordinate substitutions:

  • \( z = \rho \cos \phi \)
  • \( x^2 + y^2 = \rho^2 \sin^2 \phi \)

Substituting these into the cone equation gives:

\[ \rho \cos \phi = \sqrt{3(\rho^2 \sin^2 \phi)} \] \[ \cos \phi = \sqrt{3} \sin \phi \] \[ \tan \phi = 1/\sqrt{3} \] \[ \Rightarrow \phi = \pi/6. \]
\[ \int_{0}^{2\pi} \int_{0}^{\pi/6} \int_{0}^{2} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]

Page 7

Line integrals

Diagram showing a smooth curve labeled C in space. The curve begins at an origin point and ends at a terminal point marked with a solid dot. At an arbitrary point on the curve, three vectors are shown: a blue unit tangent vector T pointing in the direction of the curve, a green unit normal vector N pointing perpendicular to the curve, and a pink vector F representing a field vector.
Visual Description: Diagram showing a smooth curve labeled C in space. The curve begins at an origin point and ends at a terminal point marked with a solid dot. At an arbitrary point on the curve, three vectors are shown: a blue unit tangent vector T pointing in the direction of the curve, a green unit normal vector N pointing perpendicular to the curve, and a pink vector F representing a field vector.

\(C: \vec{r}(t), \quad a \le t \le b\)

Scalar Line Integral:

\[ \int_C f \, dr = \int_a^b f(\vec{r}(t)) |\vec{r}'(t)| \, dt \]

Note: \(f\) is a scalar.

Vector Line Integrals:

Tangential component:

\[ \int_C (\vec{F} \cdot \vec{T}) \, dr \]

Normal component:

\[ \int_C (\vec{F} \cdot \vec{N}) \, dr \]

Note: The dot products \((\vec{F} \cdot \vec{T})\) and \((\vec{F} \cdot \vec{N})\) are scalars.

Choice for \(\vec{N}\): The normal vector \(\vec{N}\) is defined as the vector \(90^\circ\) counter-clockwise (CCW) from the tangent vector.

Surface integral

Diagram illustrating a curved surface sheet S. At a point on the surface, two vectors are originating: a green unit normal vector N pointing vertically upwards from the surface, and a pink field vector F pointing diagonally.
Visual Description: Diagram illustrating a curved surface sheet S. At a point on the surface, two vectors are originating: a green unit normal vector N pointing vertically upwards from the surface, and a pink field vector F pointing diagonally.

\(S: \vec{r}(u,v), \quad u, v \in R \subset \mathbb{R}^2\)

Scalar Surface Integral:

\[ \underbrace{\iint_S f \, dS}_{\text{Surface Integral}} = \underbrace{\iint_R f(\vec{r}(u,v)) |\vec{r}_u \times \vec{r}_v| \, dA}_{\text{Double Integral}} \]

Vector Surface Integral (Flux):

\[ \iint_S \vec{F} \cdot d\vec{S} = \iint_S (\vec{F} \cdot \vec{N}) \, dS \]

Note: The dot product \((\vec{F} \cdot \vec{N})\) is a scalar.

\(\vec{N} = \text{unit normal pointing upwards or outwards}\)

Page 8

Line integrals: Which theorem to use

For scalar functions, we use the definition.

For vector fields, the primary question to address is: is \( C \) a closed curve in 2D?

Hand-drawn diagram of a simple closed curve in a 2D plane. The curve is labeled 'C' and encloses a region labeled 'R'. The region 'R' is filled with diagonal blue hash marks. This visual represents the setup for Green's Theorem, where a line integral around a closed boundary is related to a double integral over the interior region.
Visual Description: Hand-drawn diagram of a simple closed curve in a 2D plane. The curve is labeled 'C' and encloses a region labeled 'R'. The region 'R' is filled with diagonal blue hash marks. This visual represents the setup for Green's Theorem, where a line integral around a closed boundary is related to a double integral over the interior region.

Case 1: Yes, \( C \) is a closed curve.

Use Green's Theorem:

\[ \oint_C \vec{F} \cdot \vec{T} dr = \iint_S \text{2D curl } dA \] \[ \oint_C \vec{F} \cdot \vec{n} dr = \iint_S \text{2D div } dA \]

Case 2: No, \( C \) is not a closed curve.

Check: is \( \vec{F} = \vec{\nabla} \phi \) (conservative?)

Yes (Conservative): Use FTCLI (Fundamental Theorem of Calculus for Line Integrals)

\[ \int_C \vec{\nabla} \phi \cdot d\vec{r} = \phi(\text{end}) - \phi(\text{start}) \]

No (Not Conservative): Use the definition.

Page 9

Example: Line Integral over a Triangular Region

Let \( R \) be a triangle with vertices \( (0,0), (2,0), (1,1) \) and \( C \) be its boundary oriented counter-clockwise (CCW).

Evaluate: \[ \int_{C} \cos^3(x) \, dx + e^x \, dy \]

Line graph on a coordinate grid illustrating the triangular region R. The triangle's vertices are at (0,0), (2,0), and (1,1). The boundary C is marked with arrows showing a counter-clockwise orientation. The left boundary line is labeled y = x, and the right boundary line is labeled y = -x + 2. The interior of the triangle is shaded with horizontal lines. A pink horizontal line segment is drawn across the region at a constant y-level, spanning from the left boundary x = y to the right boundary x = 2 - y, visually representing the limits of integration for x. Points (0,0) and (2,0) are labeled on the x-axis, and the peak vertex (1,1) is also labeled.
Visual Description: Line graph on a coordinate grid illustrating the triangular region R. The triangle's vertices are at (0,0), (2,0), and (1,1). The boundary C is marked with arrows showing a counter-clockwise orientation. The left boundary line is labeled y = x, and the right boundary line is labeled y = -x + 2. The interior of the triangle is shaded with horizontal lines. A pink horizontal line segment is drawn across the region at a constant y-level, spanning from the left boundary x = y to the right boundary x = 2 - y, visually representing the limits of integration for x. Points (0,0) and (2,0) are labeled on the x-axis, and the peak vertex (1,1) is also labeled.

Since \( C \) is a simple closed curve, we can apply Green's Theorem. Define the vector field: \[ \vec{F} = \langle \cos^3 x, e^x \rangle \]

By Green's Theorem: \[ \int_{C} \vec{F} \cdot d\vec{r} = \iint_{R} (\text{curl}_z \vec{F}) \, dA \]

Calculating the integrand (\( g_x - f_y \)): \[ \frac{\partial}{\partial x}(e^x) - \frac{\partial}{\partial y}(\cos^3 x) = e^x - 0 = e^x \] So the integral becomes: \[ \iint_{R} e^x \, dA \]

To set up the double integral over region \( R \), we determine the bounds for a Type II region (integrating with respect to \( x \) first): \[ y \le x \le 2 - y \] \[ 0 \le y \le 1 \]

The integral is: \[ \int_{0}^{1} \int_{y}^{2-y} e^x \, dx \, dy \]