From Math 2220 Class 37 V1c Symmetries Can Help Flow Lines From Math 2220 Class 37 Div and Curl Why Green’s Greens Dr. Allen Back Problems Stokes and Gauss Surface of Nov. 21, 2014 Revolution Case Graph Case Surface Integrals Other Problems Surface
Symmetries Can Help From Math 2220 Class 37 V1c The most important thing you should know (e.g. for exams Symmetries Can Help and homework) is how to setup (and perhaps compute if not Flow Lines too hard) surface integrals, triple integrals, etc. Div and Curl But occasionally symmetries can save a considerable amount of Why Green’s work. Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Symmetries Can Help From Math 2220 Class 37 V1c Problem: Calculate the following surface integral: Symmetries Can Help �� y 2 − z 2 dS Flow Lines Div and Curl S Why Green’s for S the part of the paraboloid x = y 2 + z 2 with 1 ≤ x ≤ 5 . Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Symmetries Can Help In this case the symmetry of interchanging y and z is an From Math 2220 Class 37 orthogonal (and so distance/angle preserving transformation of V1c R 3 . ) (Algebraically this is the transformation Symmetries y → z , z → y , x → x . ) Can Help And this transformation preserves the region we are integrating Flow Lines over. Div and Curl Why Green’s At the Riemann sum level, for every little area ∆ S near Greens ( x , y , z ) ∈ S there is a matching area near ( x , z , y ) whose Problems Stokes and contributions at the Riemann sum level cancel. Gauss So the value of the surface integral is 0 . Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Symmetries Can Help From Math 2220 Class 37 If you use this argument on homework or an exam, please V1c 1 Describe the symmetry you are using. Symmetries Can Help 2 Mention that the region of integration is symmetric under Flow Lines this symmetry. Div and Curl 3 State clearly what cancellation or simplification results. Why Green’s Greens (In other words, don’t just say “ by symmetry” this is 0 unless Problems it really is obvious.) Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Symmetries Can Help From Math 2220 Class 37 V1c Using the same potion of a paraboloid, here are two more Symmetries examples where we can use symmetry to argue for an answer of Can Help 0 : Flow Lines Div and Curl �� S (0 , y , − z ) · ˆ n dS 1 Why Green’s �� S yz dS 2 Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Flow Lines From Math 2220 Class 37 V1c Definition: A path c : [ a , b ] → R 3 is said to be a flow line (or Symmetries integral curve) of a vector field � F ( x , y , z ) if Can Help Flow Lines c ′ ( t ) = � � F ( c ( t )) Div and Curl Why Green’s for all t ∈ [ a , b ] . Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Flow Lines From Math 2220 Class 37 V1c Symmetries Can Help Flow lines always exist at least locally because of existence and Flow Lines uniqueness theorems for ordinary differential equations. Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Flow Lines From Math 2220 Class 37 V1c Symmetries At the end of Math 2210, you studied how to use Can Help eigenvalues/eigenvectors/diagonalization/(Jordan Normal Flow Lines form) to solve linear differential equations. Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Flow Lines From Math 2220 Class 37 V1c Symmetries Can Help Show that c ( t ) = ( x 0 e t , y 0 e t ) is a flow line of � F ( x , y ) = ( x , y ) . Flow Lines Div and Curl Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Flow Lines From Math 2220 Class 37 V1c Show that Symmetries Can Help � x ( t ) � � cos t � � x 0 � − sin t c ( t ) = = Flow Lines y ( t ) sin t cos t y 0 Div and Curl Why Green’s is a flow line of � F ( x , y ) = ( − y , x ) . Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Div and Curl From Math 2220 Class 37 V1c We will discuss the meaning of these when we get to the 3d Symmetries integral theorems (8.2 and 8.4.) Can Help For a vector field � F ( x , y , z ) = ( P , Q , R ) Flow Lines div( � Div and Curl F ) is a (scalar) function; i.e. a number at each point. Why Green’s curl( � F ) is another vector field; i.e. a vector at each point. Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Div and Curl From Math 2220 Class 37 V1c div ( � F ) = P x + Q y + R z Symmetries Can Help for � F ( x , y , z ) = ( P , Q , R ). Flow Lines Div and Curl Why Green’s ( ∇ · � F ) . Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Div and Curl From Math 2220 Class 37 ˆ ˆ ˆ i j k V1c curl ( � ∂ ∂ ∂ F ) = Symmetries ∂ x ∂ y ∂ z Can Help P Q R Flow Lines for � Div and Curl F ( x , y , z ) = ( P , Q , R ). Why Green’s Greens ( ∇ × � Problems F ) . Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Div and Curl From Math 2220 Class 37 Please take a look at the table of vector analysis identities on V1c page 255. Symmetries Many are just the 1-variable product or chain rules. Can Help Flow Lines 1 ∇ ( fg ) Div and Curl 2 div ( f � F ) Why Green’s 3 curl ( f � F ) Greens Problems 4 curl ( � G × � F ) would be more complicated. Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Div and Curl From Math 2220 Class 37 V1c (There is a more complicated identity Symmetries Can Help a × ( � c ) � a · � b × � c ) = ( � a · � b − ( � b ) � � c Flow Lines Div and Curl useful in E & M, but the related curl identity (useful in some Why Green’s proofs of Stokes’ theorem is tricky to get right.) Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Div and Curl From Math 2220 Class 37 Other very important ones are V1c 1 div ( curl � F ) = 0 . Symmetries Can Help 2 curl ( ∇ f ) = 0 . Flow Lines (for C 2 functions and vector fields.) They all are consequences Div and Curl of Why Green’s f xy = f yx Greens Problems for such functions. Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Div and Curl From Math 2220 Class 37 x 2 + y 2 + z 2 . V1c � Problem: Let � r = ( x , y , z ) , and so r = � � r � = Show that a vector field of the form Symmetries Can Help Flow Lines f ( r ) � r Div and Curl Why Green’s satisfies ∇ × � F = 0 . Greens (In physics, this is called a central force field. ) Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Why Green’s From Math 2220 Class 37 Green’s theorem says that for simple closed (piecewise smooth) V1c curve C whose inside is a region R , we have Symmetries Can Help � �� ∂ Q ∂ x − ∂ P P ( x , y ) dx + Q ( x , y ) dy = ∂ y dx dy Flow Lines C R Div and Curl Why Green’s as long as the vector field � F ( x , y ) = ( P ( x , y ) , Q ( x , y )) is C 1 Greens Problems on the set R and C is given its usual “inside to the left” Stokes and orientation. Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Why Green’s From Math 2220 Class 37 V1c For a y -simple region Symmetries Can Help �� � Flow Lines − P y dy dx = P dx . Div and Curl R C = ∂ R Why Green’s is fairly easily justified. Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Why Green’s From Math 2220 Class 37 V1c For an x -simple region Symmetries Can Help �� � Flow Lines Q x dx dy = Q dy . Div and Curl R C = ∂ R Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
Why Green’s From Math 2220 Class 37 V1c So for a region that is both y -simple and x -simple we have Symmetries Green: Can Help Flow Lines � �� P ( x , y ) dx + Q ( x , y ) dy = Q y − P x dx dy . Div and Curl ∂ R R Why Green’s Greens Problems Stokes and Gauss Surface of Revolution Case Graph Case Surface Integrals Other Problems Surface
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