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MA 242.003 Day 30 - February 18, 2013 Section 11.7: Finish optimization examples Section 12.1: Double Integrals over Rectangles
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Absolute Maxima and Minima on a closed, bounded set
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Closed means “contains its boundary” Bounded means “can be enclosed in a sufficiently large circle centered on the origin”.
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(continuation of example)
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Section 12.1: Double Integrals over Rectangles The Riemann integral can be motivated by the problem of finding the AREA under the curve y = f(x) and above the x-axis for x in the interval [a,b].
![Section 12.1: Double Integrals over Rectangles The Riemann integral can be motivated by the problem of finding the AREA under the curve y = f(x) and above the x-axis for x in the interval [a,b].](http://images.slideplayer.com./32/9837662/slides/slide_10.jpg "Section 12.1: Double Integrals over Rectangles The Riemann integral can be motivated by the problem of finding the AREA under the curve y = f(x) and above the x-axis for x in the interval [a,b].")
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Section 12.1: Double Integrals over Rectangles The Riemann integral can be motivated by the problem of finding the AREA under the curve y = f(x) and above the x-axis for x in the interval [a,b].
![Section 12.1: Double Integrals over Rectangles The Riemann integral can be motivated by the problem of finding the AREA under the curve y = f(x) and above the x-axis for x in the interval [a,b].](http://images.slideplayer.com./32/9837662/slides/slide_11.jpg "Section 12.1: Double Integrals over Rectangles The Riemann integral can be motivated by the problem of finding the AREA under the curve y = f(x) and above the x-axis for x in the interval [a,b].")
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Section 12.1: Double Integrals over Rectangles The Double integral can be motivated by the problem of finding the VOLUME under the surface z = f(x,y) and above the rectangle [a,b] x [c,d].
![Section 12.1: Double Integrals over Rectangles The Double integral can be motivated by the problem of finding the VOLUME under the surface z = f(x,y) and above the rectangle [a,b] x [c,d].](http://images.slideplayer.com./32/9837662/slides/slide_12.jpg "Section 12.1: Double Integrals over Rectangles The Double integral can be motivated by the problem of finding the VOLUME under the surface z = f(x,y) and above the rectangle [a,b] x [c,d].")
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Section 12.1: Double Integrals over Rectangles The Double integral can be motivated by the problem of finding the VOLUME under the surface z = f(x,y) and above the rectangle [a,b] x [c,d].
![Section 12.1: Double Integrals over Rectangles The Double integral can be motivated by the problem of finding the VOLUME under the surface z = f(x,y) and above the rectangle [a,b] x [c,d].](http://images.slideplayer.com./32/9837662/slides/slide_13.jpg "Section 12.1: Double Integrals over Rectangles The Double integral can be motivated by the problem of finding the VOLUME under the surface z = f(x,y) and above the rectangle [a,b] x [c,d].")
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4 Steps to approximate the volume under z = f(x,y)
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Double Sum Notation:
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Double Sum Notation Example:
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(continuation of example)
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We will learn how to compute double integrals in section 11.2
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Theorem: If f(x,y) is continuous on R then f(x,y) is integrable on R.
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Applications of Double Integration
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1. Volume
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Applications of Double Integration 1. Volume 2. Average value of f(x,y) over a rectangle R
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Applications of Double Integration 1. Volume 2. Average value of f(x,y) over a rectangle R
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Applications of Double Integration 1. Volume 2. Average value of f(x,y) over a rectangle R
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Properties of Double Integrals The following properties follow from the definitions.
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Properties of Double Integrals The following properties follow from the definitions.
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Properties of Double Integrals The following properties follow from the definitions.
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Properties of Double Integrals The following properties follow from the definitions.
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Properties of Double Integrals The following properties follow from the definitions.
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(continuation of problem)
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Reading assignment: Re-read pages 438-439 on computing VOLUMES using single integration.
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