Math 208 Surface integrals and the differentials for flux integrals. n and separately. But the proof on page 889 of the formula dσ = r r du dv on page

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1 Math 08 urface integrals and the differentials for flu integrals Our tet fails to eplicitl state the formulas for n dσ, instead preferring to give formulas for n and separatel But the proof on page 88 of the formula dσ = r r du dv on page dσ u v 80 actuall shows that n dσ = ± ru rv du dv that is, r u r v du dv is a vector of length dσ and direction normal to the surface, and the onl such vectors in 3D are that and its negative! egarding n dσ as the basic formula rather than n and dσ separatel is simpler - when ou need dσ, it can be obtained as simpl n dσ, while n itself is just the unit vector in the direction of n dσ Finding n and dσ separatel and multipling them when doing a flu integral leads to computing r r twice, once in the denominator of n and once in dσ, onl to cancel these ever time! Wh do that? u v If the surface is given parametricall b u, v), then n dσ = ± r r du dv 1) Our book also fails to state the formulas for r u v n dσ when the surface is written as one variable is a function of the other two These are the most common cases used, and knowing n dσ for those cases saves several steps uppose z = f, ) with f differentiable Using and as our parameters leads to r = <,, f, ), so r r = < 0,1, f < 1,0, f = < f, f, 1, giving: Formulas for z = f, ) n dσ = ± f i + f j k) d d If the surface is part of, then ) n dσ alwas have a choice of sign which depends on the orientation of the surface involved Eg in ), + gives downward orientation,! gives upward since up/down is determined b the k coefficient) Also, be aware that the integral can, of course, be done using as indicated,, or even d d d d r dr dθ Eample 1: Find the flu of F,, z) =< 3,3, z oriented paraboloid which satisfies z + z = 0 over that portion of the upward olution: The surface equation gives us, so ince we want upward oriented, leads to: z = + f, ) = + n dσ = f i + f j k) d d = <,,1 d d,which

2 Fin dσ = < 3,3, z i<,,1 d d = 6 6 z) d d = 8 8 ) d d But is the region in the -plane where z = +, the inside of the circle of radius 3 centered at the origin, which means this integral is best done in polar coordinates We get: π 3 3 θ ) d d = 8 r ) rdrd = 8r dr dθ π θ 0 ) 4 3 = r = 34π 0 The above eample can also be done b parameterizing r in terms of r and However, as above, it is usuall simpler to set up the integral in terms of and, and then convert the integral to polar coordinates, rather than find r rθ r Note that one can rotate the roles of the variables, and get corresponding formulas: If the surface is part of = f, z), then n dσ = ± i + f j + fzk) d dz 3) and likewise = f, z) n dσ = ± f i j + f k) d dz dσ If the surface is part of, then z 4) Also note that the formulas for in these settings which are implicit in the formulas at the bottom of p 85 and top of p 86) are again just the formulas for the lengths of these vector differentials In general, if ou find ourself having trouble memorizing all of the differentials for surface integrals, memorize just the ones for flu ie the n dσ formulas), and if ou re doing a surface integral which is not a flu integral, find dσ b taking the length of the vector part of n dσ : ince n is a unit vector, n = 1, so n dσ = n dσ = dσ Thus: If the surface is given parametricall b u, v), then dσ = r r du dv 1a) r u v If the surface is part of z = f, ), then d σ = f + 1 a) f + d d

3 If the surface is part of = f, z), then d σ = 1+ f + 3a) f z d dz If the surface is part of = f, z), then d σ = f a) f z d dz Eample : Integrate G,, z) = + 4 over the surface cut from the parabolic clinder + 4z = 16 b the planes = 0, = 1, and z = 0 This is problem 14 on page 03 of the tet) olution: The surface can be written as z = 4 = f, ), so Thus n dσ = ± < f, f, 1 d d = ± < 0,, 1 d d dσ = n dσ = + d d = d d = d d z = 0 = 16 = ± 4 o 4 ) ) ) ) and on the surface, G dσ = + d d = d d = + = + = g,, z) = c The formulas given in the book for the level surface are generall harder to use than the formulas above, and the surfaces we use are almost alwas either eas to parametrize or eas to solve for one of the variables in terms of the other two, so if ou know the above formulas and know how to parametrize standard surfaces, ou re generall covered In fact, the ± g gip formula using da for n dσ as at the bottom of page 00 is precisel formula ) above if z = z p = k and the partials f and f = are calculated implicitl from the level surface equation Likewise, it is precisel formula 3) if p = i or formula 4) if p = j and the derivatives are found implicitl The onl case that is at all common where it is actuall advantageous to think in terms of n and dσ separatel is in the special case where flu can be found geometricall If Fin = scalar component of F in the direction of n) = c 1 is constant on the surface note F can alwas be written as a vector parallel to n plus a vector perpendicular to n, and this sas the length of the part parallel to n stas the same on the entire surface) and the surface area of is

4 known, then Fin dσ = c1 urface area of ) Eample 3: Find the flu of sphere F,, z) = < 4,4,4z + + z =, oriented awa from the origin across the first octant portion of the olution: Note that F F = 4 <,, z = z On this sphere, that means = 4 = 1 But also, F points directl awa from the origin, which is the same direction as the outward unit normal n to the sphere at that point Thus at each point on the sphere, Fin = F n cos0 ) = F = 1 = c 1 since n is a unit normal) Also, the surface area of a sphere of radius r is 4 r π eas to remember since it s the derivative of the volume), or in this case 36B, which means the surface area of the first octant portion is 1 π = 54π 36π = 8 π Thus the flu is Flu integral problems 1 Find the flu of over the upward oriented portion of F,, z) = < z,0,4 z = F,, z ) = < 3,1, that is defined b and Find the flu of over the portion of the surface + z = z 4 that has, and is oriented toward decreasing F,, z) = <,, z z = = z = z 3 Find the flu of over the finite piece of 4 = z bounded b,, and, if the orientation is awa from ou as viewed from a point on the negative -ais 4 Find the flu of each of the following through the portion of the clinder + z = 5 with 1, if the surface is oriented toward the -ais ie, the clinder is oriented inward): a) F,, z) = < 0,,z b) F,, z) = < 3 z, 4, 4z

5 Answers: 1!40 7 3!384 4 a)!300b b) 600B