{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture12 - Lecture 11 Double integrals Lecture 11 Double...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Lecture 11: Double integrals May 28, 2009 Lecture 11: Double integrals
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Objectives 1 Interpret integrals as volumes. 2 Define average value of a function. 3 Use partial integration to evaluate double integrals on rectangles. 4 Use partial integration to evaluate double integrals on more general regions. Lecture 11: Double integrals
Background image of page 2
Integration in one dimension. Z b a f ( x ) dx = lim N →∞ N X i =1 f ( x * i x ! When f ( x ) > 0, the integral is interpreted as the area between the graph of f ( x ) and the x -axis between the points a and b . Lecture 11: Double integrals
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Integration in two dimensions ZZ R f ( x , y ) dA = lim m , n →∞ n X i =1 m X j =1 f ( x * ij , y * ij A R = { ( x , y ) | a x b , c y d } When f ( x , y ) > 0, the integral is interpreted as the volume between the graph of f ( x , y ) and the xy -plane, bounded by the rectangle R . Lecture 11: Double integrals
Background image of page 4
Class exercise ZZ R (5 - x ) dA R = { ( x , y ) | 0 x 5 , 0 y 3 } Lecture 11: Double integrals
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Class exercise ZZ R (5 - x ) dA R = { ( x , y ) | 0 x 5 , 0 y 3 } Answer: 75 2 Lecture 11: Double integrals
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}