Multiple Integrals

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Representing Functions as Power Series

Functions as Power Series
Derivatives and Integrals of Power Series
Applications and Examples

Taylor and Maclaurin Series

The Formula for Taylor Series
Taylor Series for Common Functions
Adding, Multiplying, and Dividing Power Series
Miscellaneous Useful Facts

Applications of Taylor Polynomials

Taylor Polynomials
When Functions Are Equal to Their Taylor Series
When a Function Does Not Equal Its Taylor Series
Other Uses of Taylor Polynomials

Partial Derivatives

Visualizing Functions in 3 Dimensions
Definitions and Examples
An Example from DNA
Geometry of Partial Derivatives
Higher Order Derivatives
Differentials and Taylor Expansions

Multiple Integrals

Background
What is a Double Integral?
Volumes as Double Integrals

Iterated Integrals over Rectangles

How To Compute Iterated Integrals
Examples of Iterated Integrals
Cavalieri's Principle
Fubini's Theorem
Summary and an Important Example

Double Integrals over General Regions

Type I and Type II regions
Examples 1-4
Examples 5-7
Order of Integration

Volumes as Double Integrals

Double integrals can be viewed as volumes in the same way that regular integrals can be viewed as areas.

Previously, we defined the area under a curve to be an integral, and now we define the volume under a surface to be a double integral:

Definition: Let

$z=f(x,\,y) \ge 0$ and let

$R$ be a rectangle in the

$xy$ -plane. The double integral

$\iint_R\, f(x,y)\, dx \,dy$ of

$f$ over

$R$ is the volume of the solid under the graph of $f$ and above $R$ .

We can compute volumes by computing double integrals. Here , we compute double integrals by using what we already know about 3-dimensional geometry.

Example 1: Compute the integral

$\iint_R \frac{4}{3}(5-x) \,dA$ , where

$R$ is the rectangle

$[2,\,5]\,\times\,[1,\, 3]$ in the

$xy$ -plane.

Solution 1: The integral is equal to the volume of the solid shown here, which we call

$W$ .

$W$ is the region above

$R$ on the

$xy$ -plane and under the (tilted) plane

$z=\frac{4}{3}(5-x)$ . To find its volume, take a vertical slice for a fixed

$y$ with

$1\le y\le 3$ . The slice of the solid on the vertical

$xz$ -plane is the same triangle for each

$y$ . We can see from the diagram that the slice has area

$\frac{1}{2}\times \hbox{base}\times \hbox{height} \ =\frac12\cdot 3\cdot 4=6$ Then

$W$ has volume

$V$ which is the area of the triangle times the length of the side. So

$V=6\cdot 2 = 12$ . We get

$\iint_R \frac{4}{3}(5-x)\, dA = \hbox{ volume of $W$ } =V= 12.$

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Differential Equations

Models of Growth

Infinite Sequences

Infinite Series

Integral Test

Comparison Tests

Convergence of Series with Negative Terms

The Ratio and Root Tests

Strategies for testing Series

Power Series

Representing Functions as Power Series

Taylor and Maclaurin Series

Applications of Taylor Polynomials

Partial Derivatives

Multiple Integrals

Iterated Integrals over Rectangles

Double Integrals over General Regions

Volumes as Double Integrals