MIT OpenCourseWare http ocw mit edu 18 02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use visit http ocw mit edu terms 18 02 Lecture 24 Tue Nov 6 2007 Simply connected regions slightly di erent from the actual notations used Recall Green s theorem if C is a closed curve around R counterclockwise then line integrals can be expressed as double integrals F d r curl F dA F n ds div F dA C R C R where curl M Nj Nx My div P Qj Px Qy For Green s theorem to hold F must be de ned on the entire region R enclosed by C y xj C unit circle counterclockwise then curl F Example same as in pset F 2 x y2 y x 2 2 0 So if we look at both sides of Green s theorem 2 x x y y x y 2 F d r 2 from pset curlF dA 0 dA 0 C R R The problem is that R includes 0 where F is not de ned De nition a region R in the plane is simply connected if given any closed curve in R its interior region is entirely contained in R Examples shown So Green s theorem applies safely when the domain in which F is de ned and di erentiable is simply connected then we automatically know that if F is de ned on C then it s also de ned in the region bounded by C In the above example can t apply Green to the unit circle because the domain of de nition of F is not simply connected Still we can apply Green s theorem to an annulus picture shown of a curve C unit circle counterclockwise segment along x axis small circle around origin clockwise back to the unit circle allong the x axis enclosing an annulus R Then Green applies and says C F d r R 0 dA 0 but line integral simpli es to C C C2 where C unit circle C2 small circle origin so line integral is actually the same on C and C2 or any other curve encircling the origin Review for Exam 3 2 main objects double integrals and line integrals Must know how to set up and evaluate Double integrals drawing picture of region taking slices to set up the iterated integral Also in polar coordinates with dA r dr d see e g Problem 2 not done Remember mass centroid moment of inertia For evaluation need to know usual basic integrals e g dx x integration by substitution e g 1 2 t dt du setting u 1 t2 Don t need to know complicated trigonometric 2 2 1 t 0 1 4 u integrals e g cos d integration by parts Change of variables recall method u v ux uy 1 Jacobian Its absolute value gives ratio between du dv and dx dy x y cx vy 2 express integrand in terms of u v 1 2 3 set up bounds in uv coordinates by drawing picture The actual example on the test will be reasonably simple constant bounds or circle in uv coords Line integrals C F d r C F T ds C M dx N dy To evaluate express both x y in terms of a single parameter and substitute Special case gradient elds Recall F is conservative F d r is path independent F is the gradient of some potential f curl F 0 i e Nx My If this is the case then we can look for a potential using one of the two methods antiderivatives or line integral and we can then use the FTC to avoid calculating the line integral cf Problem 3 Flux F n ds Q dx P dy Geometric interpretation C C Green s theorem in both forms already written at beginning of lecture 18 02 Lecture 25 Fri Nov 9 2007 Handouts Exam 3 solutions Triple integrals f dV dV volume element R Example 1 region between paraboloids z x2 y 2 and z 4 x2 y 2 picture drawn e g 4 x2 y2 volume of this region 1 dV dz dy dx R x2 y 2 To set up bounds 1 for xed x y nd bounds for z here lower limit is z x2 y 2 upper limit is z 4 x2 y 2 2 nd the shadow of R onto the xy plane i e set of values of x y above which region lies Here R is widest at intersection of paraboloids which is in plane z 2 general method for which x y is z on top surface z on bottom surface Answer when 4 x2 y 2 x2 y 2 2 2 i e x y 2 So we integrate over a disk of radius 2 in the xy plane By usual method to set up double integrals we nally get 2 2 x2 4 x2 y2 V dz dy dx 2 2 x2 x2 y 2 Evaluation would be easier if we used polar coordinates x r cos y r sin x2 y 2 r2 then 2 2 4 r2 V dz r dr d 0 0 r2 evaluation easy not done Cylindrical coordinates r z x r cos y r sin r measures distance from z axis measures angle from xz plane picture shown Cylinder of radius a centered on z axis is r a drawn 0 is a vertical half plane not drawn Volume element in rect coords dV dx dy dz in cylindrical coords dV r dr d dz In both cases this is justi ed by considering a small box with height z and base area A then volume is V A z Applications Mass M R dV 3 1 Average value of f over R f V ol 1 f dV weighted average f f dV M ass R R 1 In particular center of mass x y z where x x dV M ass R Note can sometimes avoid calculation using symmetry e g in above example x y 0 Moment of inertia around an axis I distance from axis 2 dV R 2 2 About z axis Iz r dV x y 2 dV consistent with I0 in 2D case R R 2 2 Similarly about x and y axes Ix y z dV Iy x2 z 2 dV R R setting z 0 this is consistent with previous de nitions of Ix and Iy for plane regions Example 2 moment of inertia Iz of solid cone between z ar and z b 1 picture drawn b 2 z a b5 2 2 Iz r dV r r dr d dz 10a4 R 0 0 0 I explained how to nd bounds in order dr d dz rst we x z then slice for given z is the disk bounded by r z a the rst slice is z 0 the last one is z b Example 3 volume of region where z 1 y and x2 y 2 z 2 1 Pictures drawn in space slice by yz …
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