Calculus III MATH 20550
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This 0 page Class Notes was uploaded by Briana Cassin on Sunday November 1, 2015. The Class Notes belongs to MATH 20550 at University of Notre Dame taught by Laurence Taylor in Fall. Since its upload, it has received 33 views. For similar materials see /class/232698/math-20550-university-of-notre-dame in Mathematics (M) at University of Notre Dame.
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Date Created: 11/01/15
There are a couple of types of surfaces that can always be pararneterized 1 GRAPHS If the surface is z ay for all Ly in the region D7 then take z u7 y w and z fuw for u7w E D 11 Upper hemisphere The graph is z xaz 7 2 7 y with the region D being x2 y2 g 12 One can of course take pieces ofthis The sphere of radius a in the rst octant is z xaz 7 x2 7 y with D being x2 y2 g 127 z 2 07 y 2 0 The part of the upper hernisphere lying outside the 2 cylinder 2 y2 6 Here is D 12 A plane If the surface is ax by 02 d 0 one of 17 b or c is non zero Solve for the corresponding variable For exarnple7 if c 31 0 then 2 ow By y 2 SURFACES OF REVOLUTION Given a pararneterized curve in the zy plane7 say z Mt7 y gt7 a g t g b7 the corresponding surface of revolution is given by rotating this curve about say the y axis If you pick a point in the xy plane and rotate about the y axis7 the y coordinate stays constant and the in the x2 coordinates you see a circle of radius x Hence if you use 6 to be the rotation coordinate7 the surface becornes z t cos67 y gt7 z t sin67 a g t g b7 0 g 6 g 27139 21 The torus Take a circle of radius b centered at 10 in the zy plane Assume 0 lt b lt a Rotate this circle about the y axis You get atorus The pararneterization is z ab cost cos6 y bsint7 z a bcost sin67 0 g t g 27f7 0 g 6 g 27139 3 SURFACE INTEGRALS Sfz7dS where S is some surface and f is a function de ned on that surface The de nition says cut up the surface into lots of little pieces7 evaluate the function at a point in a piece7 multiply by the area of the piece and add Then take a limit Here is a picture of a torus suitably chopped up We rst reduce the surface integral to a planar double integral 1 Pararneterize the surface z u7w7 y yuw7 z zuw7 u7w E D or as Fuw7 u7w E D 2 Calculate FM and 67 Fw Then calculate the normal vector IV FM gtlt 77v 3 fz7dS Karma 17 dA S D 4 Do the planar double integral by any of the methods available to you Here are formulae for IV and hence dS7 in the two cases discussed above 31 Graphs Fxy ltzyfxygt so Fm lt1707fwgt7 Fy lt0717fygt and a i j k Ndet1 0 fm i7fm7jfyk1lt7fm7fy1gt 0 1 fy sodS1f f3 dA 2 32 Surfaces of revolution For a curve in the xy plane rotated about the y axis 77t7 6 cos 6ytzt sin 6gt so Ft ltx cos 67y 7 x sin 6gt7 F9 lt7x sin 67 07 xcos 6gt and i k ipy cos 6 l det x cos 6 y x sin 6 ijzx cos2 6 zx sin2 6 ltxy cos 67 pp xy sin 6gt 7p sin6 0 pcos6 kaySin 9 Hence dS day2 z2x 2 dA y2 x 2 dA This gives a famous theorem known to the Greek mathematician Pappus of Alexandria around 320 AD Let C be a curve in the plane to the right of the y axis and let S be the solid obtained by rotating 0 around the y axis Suppose C is parameterized by xt7 gt7 a g t g h Then 27r The areaofS1dSxqy 2z 2 dA zdsd627TMy S R 0 C We can write My length of C where i is the p coordinate of the centroid This yields the result as given by Pappus in English Given a connected curve C lying in a plane and a line L in that plane which does not intersect the curve the area of the surface obtained by rotating 0 around L is equal to the product of the length of C and the distance travelled by the centroid of C under the rotation Let f be a function of a vector variable of 71 variables x 17 7n f9617 7 W The function is dz erentz39able at the point 6 117 7an provided the partial derivatives exist 239 1 n and the following formula holds Let Ax differentiable we additionally require that 5f a 69510 A17 Azn For f to be lirn Amati m A95 7 ex Sigma lml Let Af6A f5i A 7 f6 Then a mom 2 673mm i1 provided the partials exist Dz erentmble rneans roughly that the right hand side is the best linear approximation to Af that there is If the partials are continuous in a small ball around 6 then f is differential at the point 55 The graditmt of a function ff is an n vector provided 2 is an n vector lf 2 17 7x then 6f 6f WW 67 ltzgt The formulas above can be expressed more succinctly using the gradient and the dot product Af6Ax Vf6oAx The chain rule can be summarized as follows If xi 7 239 17 n where f 13 7ztm then let gt1 7tm fz1f Fix all the ts but one7 say tk Let g 1037 712 77 nal 71 71 where t is in the kth position Then am as t ittm tm m0 mm gt Wm gt 39 n a EOE 7tm Vf Tk Tka One class of problems studies how ff varies as i does The derivative of f at the point 6 in the direction of i7 is 1 Dif5 W Vf 3917 The direction of maximal increase of the function f at the point 6 is the direction of Vf6 and the value of this rate of increase is lVf6l The direction of maximal decrease is the direction of 7Vfd and the value is ilV d lf Vf6 6 there are no such directions A second class of problems studies a xed surface ff c To complicate matters still further these two kinds of problems overlap One way to study how ff changes is to study the level surfaces ff c If a point i is xed on ff c then fEi 0 Suppose 770 is a curve with Ft0 5i and suppose moreover that 770 lies on ff c This means fFt c By the Chain Rule 0 We o mo so the tangent vector to the curve is normal to the gradient The tangent line plane is the object containing the point 6 and perpendicular to Vf lf 6 a1 an and if Vf6 b1 bn then the equation for the tangent object is biWi 01bnnan 0 or Vf6i 0276 0 6 If 616 31 0 and all the partials are continuous near 55 then there is a theorem called the s Implicit Function Theorem which says there exists a differentiable function 9x1 xn1 de ned in a neighborhood of a1 an1 such that flt17quot397n71791739 7n71 C One can ask which direction on c is the direction of maximal increase on the curve surface This is the same question as asking for the direction of maximal increase of g and so we need to compute the gradient of 9 By the Chain Rule for each i 1 g i g n 71 a C99 0Vfxolt0 1 067igt where the 1 is in the ith position Hence 5f 39 5 3 7 3f 67 and so 5f 5f
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