ADEFGDALKJHA r in .NET

Drawing QR Code ISO/IEC18004 in .NET ADEFGDALKJHA r

ADEFGDALKJHA r
QR Code ISO/IEC18004 Reader In Visual Studio .NET
Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in .NET applications.
QR Code ISO/IEC18004 Drawer In .NET
Using Barcode generator for .NET Control to generate, create Denso QR Bar Code image in VS .NET applications.
Fig. 10-11
QR-Code Recognizer In .NET Framework
Using Barcode recognizer for .NET Control to read, scan read, scan image in .NET applications.
Barcode Maker In .NET Framework
Using Barcode creator for .NET Control to generate, create barcode image in .NET framework applications.
But the integral on the left, leaving out the integrand, is equal to since
Barcode Reader In Visual Studio .NET
Using Barcode recognizer for .NET framework Control to read, scan read, scan image in .NET framework applications.
Paint QR Code In C#
Using Barcode creation for Visual Studio .NET Control to generate, create QR Code ISO/IEC18004 image in Visual Studio .NET applications.
ALKJHA
QR Code JIS X 0510 Creator In .NET
Using Barcode generator for ASP.NET Control to generate, create QR Code image in ASP.NET applications.
QR Code 2d Barcode Maker In VB.NET
Using Barcode printer for .NET framework Control to generate, create QR image in Visual Studio .NET applications.
DEFGD
1D Barcode Creator In VS .NET
Using Barcode maker for Visual Studio .NET Control to generate, create 1D image in Visual Studio .NET applications.
Encode Bar Code In .NET Framework
Using Barcode encoder for .NET Control to generate, create barcode image in .NET applications.
ALKJHA
Encoding Universal Product Code Version A In VS .NET
Using Barcode creation for .NET framework Control to generate, create Universal Product Code version A image in Visual Studio .NET applications.
UCC - 12 Creation In .NET Framework
Using Barcode drawer for .NET Control to generate, create GTIN - 12 image in Visual Studio .NET applications.
DEFGD
GTIN - 12 Recognizer In Visual Basic .NET
Using Barcode reader for Visual Studio .NET Control to read, scan read, scan image in .NET framework applications.
UPCA Scanner In .NET Framework
Using Barcode recognizer for .NET framework Control to read, scan read, scan image in .NET applications.
Thus, if C1 is the curve ALKJHA, C2 is the curve DEFGD and C is the boundary of r consisting of C1 and C2 (traversed in the positive directions), then and so P dx Q dy
Code 128A Printer In Java
Using Barcode encoder for BIRT reports Control to generate, create USS Code 128 image in Eclipse BIRT applications.
UCC.EAN - 128 Generator In .NET Framework
Using Barcode creator for ASP.NET Control to generate, create EAN128 image in ASP.NET applications.

Painting UPC Code In Java
Using Barcode drawer for Java Control to generate, create UPC Code image in Java applications.
Print ECC200 In Java
Using Barcode printer for BIRT Control to generate, create ECC200 image in BIRT applications.
@Q @P dx dy @x @y
Drawing Bar Code In Java
Using Barcode printer for BIRT Control to generate, create bar code image in BIRT applications.
GS1 - 12 Creator In Java
Using Barcode printer for BIRT reports Control to generate, create GTIN - 12 image in BIRT applications.
CHAP. 10]
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
INDEPENDENCE OF THE PATH 10.11. Let P x; y and Q x; y be continuous and have continuous rst partial derivatives at each point of Prove that a necessary and su cient condition that a simply connected region r. P dx Q dy 0 around every closed path C in r is that @P=@y @Q=@x identically in r.
Su ciency.
Suppose @P=@y @Q=@x.
Then by Green s theorem,   @Q @P dx dy 0 P dx Q dy @x @y C
where r is the region bounded by C. Necessity. Suppose r.
P dx Q dy 0 around every closed path C in r and that @P=@y 6 @Q=@x at some point of
In particular, suppose @P=@y @Q=@x > 0 at the point x0 ; y0 . By hypothesis @P=@y and @Q=@x are continuous in r, so that there must be some region  containing x0 ; y0 as an interior point for which @P=@y @Q=@x > 0. If is the boundary of , then by Green s theorem   @Q @P P dx Q dy dx dy > 0 @x @y
contradicting the hypothesis that P dx Q dy 0 for all closed curves in r. Thus @Q=@x @P=@y cannot be positive. Similarly, we can show that @Q=@x @P=@y cannot be negative, and it follows that it must be identically zero, i.e., @P=@y @Q=@x identically in r.
10.12. Let P and Q be de ned as in Problem 10.11. Prove that a B necessary and su cient condition that P dx Q dy be independent of the path in r joining points A and B is that @P=@y @Q=@x identically in r.
Su ciency. If @P=@y @Q=@x, then by Problem 10.11, P dx Q dy 0
ADBEA
D C1 A C2 E
Fig. 10-12
(see Fig. 10-12).
From this, omitting for brevity the integrand P dx Q dy, we have 0; and so
ADB BEA ADB BEA AEB C1 C2
i.e., the integral is independent of the path. Necessity. If the integral is independent of the path, then for all paths C1 and C2 in r we have ; and 0
C1 C2 ADB AEB ADBEA
From this it follows that the line integral around any closed path in r is zero, and hence by Problem 10.11 that @P=@y @Q=@x.
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
[CHAP. 10
10.13. Let P and Q be as in Problem 10.11. (a) Prove that a necessary and su cient condition that P dx Q dy be an exact di erential of a function  x; y is that @P=@y @Q=@x. B B P dx Q dy d  B  A where A and B are any two (b) Show that in such case A A points.
(a) Necessity. @ @ dx dy, an exact di erential, then (1) @=@x P, (2) @=@y 0. @x @y Thus, by di erentiating (1) and (2) with respect to y and x, respectively, @P=@y @Q=@x since we are assuming continuity of the partial derivatives. If P dx Q dy d Su ciency. By Problem 10.12, if @P=@y @Q=@x, then P dx Q dy is independent of the path joining two points. In particular, let the two points be a; b and x; y and de ne x;y  x; y Then  x x; y  x; y
x;y
P dx Q dy
a;b
x x;y
x;y P dx Q dy
a;b
P dx Q dy
a;b x x;y
P dx Q dy
Since the last integral is independent of the path joining x; y and x x; y , we can choose the path to be a straight line joining these points (see Fig. 10-13) so that dy 0. Then by the mean value theorem for integrals,  x x; y  x; y 1 x x x x;y
x;y
P dx P x  x; y
0<<1
Taking the limit as x ! 0, we have @=@x P. Similarly we can show that @=@y Q. Thus it follows that P dx Q dy
@ @ dx dy d: @x @y
(x, y)
(x + Dx, y)
(a, b) x
Fig. 10-13
CHAP. 10]
LINE INTEGRALS, SURFACE INTEGRALS, AND INTEGRAL THEOREMS
(b) Let A x1 ; y1 ; B x2 ; y2 .
From part (a), x;y P dx Q dy  x; y
a;b
Then omitting the integrand P dx Q dy, we have B x2 ;y2 x2 ;y2 x1 ;y1  x2 ; y2  x1 ; y1  B  A
Copyright © OnBarcode.com . All rights reserved.