Slide_11 - MA1104 Multivariable Calculus Lecture 11 Dr KU...

Info iconThis preview shows pages 1–17. Sign up to view the full content.

View Full Document Right Arrow Icon
MA1104 Multivariable Calculus Lecture 11 Dr KU Cheng Yeaw Thursday Feb 19, 2009
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
Recap Last lecture . .... we defined double integral over a rectangle region R via Fubini’s Theorem, we can evaluate double integral over a rectangle region using iterated integral we extend our definition of double integral to that that over a more general region D .
Background image of page 2
Overview In this lecture, we learn how to evaluate double integral over Type I and Type II region we learn how to evaluate double integral over polar rectangle.
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
Double Integral over General Region Recall how we have defined double integral of f over a general region D : We first define a new function F ( x,y ) = ± f ( x,y ) if ( x,y ) D 0 if ( x,y ) R - D If F is integrable over R , then we define the double integral of f over D by ZZ D f ( x,y ) dA = ZZ R F ( x,y ) dA.
Background image of page 4
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
In general, F ( x,y ) may not be integrable. However, we can avoid this problem given some mild conditions. Double Integral Over Bounded Region Let D be a domain whose boundary is a simple, closed, piecewise smooth curve. If f ( x,y ) is continuous on D , then ZZ D f ( x,y ) dA = ZZ R F ( x,y ) dA exists.
Background image of page 6
Now, we shall focus on two particular types of domain D which is a region between two curves. They satisfy the assumptions in the previous theorem which ensures the integrability of F ( x,y ) . We call these domain to be of Type I and Type II .
Background image of page 7

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

View Full Document Right Arrow Icon
Type I Domain Type I Domain A plane region D is said to be of Type I if it lies between the graphs of two continuous functions of x , that is, D = { ( x,y ) : a x b,g 1 ( x ) y g 2 ( x ) } where g 1 ( x ) and g 2 ( x ) are continuous on [ a,b ] .
Background image of page 8
Some examples of Type I region:
Background image of page 9

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

View Full Document Right Arrow Icon
Background image of page 10
How do we compute the integral of f ( x,y ) over such D ? Integral over Type I Domain If f is continuous on a Type I domain D such that D = { ( x,y ) : a x b,g 1 ( x ) y g 2 ( x ) } then ZZ D f ( x,y ) dA = Z b a Z g 2 ( x ) g 1 ( x ) f ( x,y ) dy dx.
Background image of page 11

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

View Full Document Right Arrow Icon
Observe that the expression on the right-hand side of ZZ D f ( x,y ) dA = Z b a Z g 2 ( x ) g 1 ( x ) f ( x,y ) dy dx is an iterated integral similar to the ones we have for rectangle region, except that in the inner integral we regard x as being constant not only in f ( x,y ) but also in the limits of the integration , g 1 ( x ) and g 2 ( x ) .
Background image of page 12
Lets prove the theorem, well, actually, we are not able to give a complete proof . .. Proof. By definition, we need to choose a rectangle R = [ a,b ] × [ c,d ] that contains D and set F ( x,y ) = ± f ( x,y ) if ( x,y ) D 0 if ( x,y ) R - D
Background image of page 13

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

View Full Document Right Arrow Icon
Background image of page 14
Then, ZZ D f ( x,y ) dA = ZZ R F ( x,y ) dA = Z b a Z d c F ( x,y ) dy dx. Remark. Although F ( x,y ) need not be continuous, the use of Fubini’s Theorem in the second equation above can be justified (however, we will not supply the details here). In particular, R d c F ( x,y ) dy exists and is a continuous function of x .
Background image of page 15

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

View Full Document Right Arrow Icon
Observe that F ( x,y ) = 0 if y < g 1 ( x ) or y > g 2 ( x ) because then ( x,y ) lies outside D . Therefore, Z d c F (
Background image of page 16
Image of page 17
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 71

Slide_11 - MA1104 Multivariable Calculus Lecture 11 Dr KU...

This preview shows document pages 1 - 17. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online