Double Integrals in Polar Coordinates: Evaluating Integrals and Finding Areas and Volumes, Study notes of Mathematics

A portion of lecture notes from a university mathematics course, math 114-004, taught at the university of pennsylvania during the fall 2009 semester. The notes cover the topic of double integrals in polar coordinates, with examples on evaluating integrals, finding areas, and calculating volumes. The professor, tong zhu, explains how to use polar coordinates to transform the integrals and provides geometric interpretations of the differential area da.

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Pre 2010

Uploaded on 03/28/2010

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Math 114-004, Fall 2009
Tong Zhu
Department of Mathematics
University of Pennsylvania
November 12, 2009
Tong Zhu Math 114-004, Fall 2009
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Math 114-004, Fall 2009

Tong Zhu

Department of Mathematics University of Pennsylvania

November 12, 2009

Double Integrals in Polar Coordinates

Double Integrals in Polar Coordinates

Example 1 Evaluate

D xydA^ where^ D^ is the disk centered at the origin with radius 3.

Polar Coordinate System

Polar Coordinate System

x = r cos θ y = r sin θ

How can we use polar coordinates to evaluate ∫ ∫

D

f (x, y )dA?

How can we use polar coordinates to evaluate ∫ ∫

D

f (x, y )dA?

This is in fact a special case of change of variables in multiple integrals, which will be shown in later section.

Geometrically, what is dA in the polar coordinates?

How can we use polar coordinates to evaluate ∫ ∫

D

f (x, y )dA?

This is in fact a special case of change of variables in multiple integrals, which will be shown in later section.

Geometrically, what is dA in the polar coordinates?

How can we use polar coordinates to evaluate ∫ ∫

D

f (x, y )dA?

This is in fact a special case of change of variables in multiple integrals, which will be shown in later section.

Geometrically, what is dA in the polar coordinates?

dA = rdrdθ ⇒

D

f (x, y )dA =

D

f (r cos θ, r sin θ)r drdθ

Revisit Example 1 Evaluate

D xydA^ where^ D^ is the disk centered at the origin with radius 3.

Example 3 Find the volume of the solid under the cone z =

x^2 + y 2 and above the disk x^2 + y 2 ≤ 4.