The differentials for flux integralsMath 208Our text fails to explicitly state the formula for when the surface is written as one variable is adσnfunction of the other two. This is by far the most common case used, and knowing savesdσnseveral steps, so it should be known. Most often, z is a function of x and y. In that case we get:If the surface is part of , then .( , )z f x y=( )x yd f f dy dxσ=± + −n i j kThe choice of sign depends on the orientation of the surface involved, + giving downward orientation,! giving upward (since up/down is determined by the k coefficient). Also, be aware that the integralcan, of course, be done as indicated, , or even . But treating as ady dxdxdyr dr dθdσncombination and taking the dot product with the vector field is almost always simpler than findingF and separately, calculating , and then multiplying by . For example:ndσF nidσExample: Find the flux of over that portion of the upward oriented( , , ) 3 ,3 , 2x y z x y z=< − >Fparaboloid which satisfies . 2 20x y z+ − =9z≤Solution: The surface equation, solved for z, gives us so we need2 2z x y= +. Since we want upward oriented,2 2( , )f x y x y= +,which leads to( ) 2 , 2 ,1x yd f f dy dx x y dy dxσ=− + − =< − − >n i j k2 22 23 ,3 , 2 2 , 2 ,1( 6 6 2 )( 8 8 )xyxyxyS RRRd x y z x y dy dxx y z dydxx y dydxσ= < − > < − − >= − − −= − −∫∫ ∫∫∫∫∫∫F ni iBut Rxy is the region in the xy-plane where , which means this integral is best done2 29z x y= + ≤in polar coordinates. We get:()2 32 2 30 032400( 8 8 ) 82 324xyRx y dydx r dr drππθθ π− − = − = − = −  ∫∫ ∫ ∫Note that one can “rotate” the roles of the variables, and get corresponding formulas: If the surface is part of , then .( , )x f y z=( )y zd f f dy dzσ=± − + +n i j kand likewiseIf the surface is part of , then .( , )y f x z=( )x zd f f dxdzσ=± − +n i j kAlso note that the formulas for in these settings (which are in the text at the bottom of p. 895 anddσtop of p. 896) are just the formulas for the lengths of these vector differentials. In general, if you findyourself having trouble memorizing all of the differentials for surface integrals, memorize the ones for, and if you’re doing a surface integral which is not a flux integral, find by taking thedσndσlength of the vector part. The formula for is generally simpler to memorize and use than thedσnformulas for and done separately. (When done separately, some extra square roots arise,ndσwhich eventually cancel.) For we mainly use the above cases and the parametric case, wheredσnagain the combined differential is simpler to use than and separately, but it is not givenndσexplicitly in the text:If the surface is given parametrically by , then .( , )u vr( )u vd du dvσ=± ×n r rThe formulas given in the book for the level surface are generally harder to use than( , , )g x y z c=the formulas above, and the surfaces we give you are always either easy to parametrize or easy to solvefor one of the variables in terms of the other two, so if you know the above formulas and know how toparametrize standard surfaces, you’re generally covered.Exercises:1. Find the flux of over the upward oriented portion of2( , , ) 2 ,0,4x y z z y= < >F that is defined by and .22 3z x y= + +1 3x− ≤ ≤1 4y≤ ≤2. Find the flux of over the portion of the surface ( , , ) 3,1, 2x y z= < − >F2 2y z x+ =that has , and is oriented toward decreasing y.2x−2 ≤ ≤2z− ≤ ≤ 43. Find the flux of over the finite piece of bounded by( , , ) 2 , ,x y z x y z= < >F2 4x y z=, , and , if the orientation is away from you as viewed from a point2z= −y z= −2y z=on the negative x-axis.Answers:1.!2402. 723.