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 V10 The Divergence Theorem 1 Introduction statement of the theorem The divergence theorem is about closed surfaces so let s start there By a closed surface S we will mean a surface consisting of one connected piece which doesn t intersect itself and which completely encloses a single finite region D of space called its interior The closed surface S is then said to be the boundary of D we include S in D A sphere cube and torus an inflated bicycle inner tube are all examples of closed surfaces On the other hand these are not closed surfaces a plane a sphere with one point removed a tin can whose cross section looks like a figure 8 it intersects itself an infinite cylinder hn A closed surface always has two sides and it has a natural positive direction the one for which n points away from the interior i e points toward the outside We shall always understand that the closed surface has been oriented this way unless otherwise specified We now generalize to 3 space the normal form of Green s theorem Section V4 Definition Let F x y z Mi N j P k be a vector field differentiable in some region D By the divergence of F we mean the scalar function div F of three variables defined in D by The divergence theorem Let S be a positively oriented closed surface with interior D and let F be a vector field continuously differentiable in a domain contatining D Then We write dV on the right side rather than dxdy dz since the triple integral is often calculated in other coordinate systems particularly spherical coordinates The theorem is sometimes called Gauss theorem Physically the divergence theorem is interpreted just like the normal form for Green s theorem Think of F as a three dimensional flow field Look first at the left side of 2 The surface integral represents the mass transport rate across the closed surface S with flow out of S considered as positive flow into S as negative Look now at the right side of 2 In what follows we will show that the value of div F at x y z can be interpreted as the source rate at x y z the rate at which fluid is being added to the flow at this point Negative rate means fluid is being removed from the flow The integral on the right of 2 thus represents the source rate for D So what the divergence theorem says is 3 Aux across S source rate for D i e the net flow outward across S is the same as the rate at which fluid is being produced or added to the flow inside S V VECTOR INTEGRAL CALCLUS 2 To complete the argument for 3 we still have to show that div F source rate a t x y z n To see this let Po xo yo zo be a point inside the region D where F is defined To simplify we denote by div F o d M d etc the value of these functions at Po Consider a little rectangular box with edges Ax Ay Az parallel to the coordinate axes and one corner at Po We take n to be always pointing outwards as usual thus on top of the box n k but on the bottom face n k A A 7 M I A X The flux across the top face in the n direction is approximately F xo yo zo Az k AxAy P xo yo zo Az AxAy while the flux across the bottom face in the n direction is approximately F x0 Yo zo k AXAY P xo Yo zo AXAY So the net flux across the two faces combined is approximately P xo YO zo Az x oYO ZO AxAy g AXAYAZ Since the difference quotient is approximately equal to the partial derivative we get the first line below the reasoning for the following two lines is analogous net Aux across two side faces AxAyAz net Aux across top and bottom o x y z 0 net Aux across front and back g A A AZ Adding up these three net fluxes and using 3 we see that source rate for box net Aux across faces of box Using this we get the interpretation for div F we are seeking source rate a t Po lim box to source rate for box div F o volume of box Example 1 Verify the theorem when F x i Solution For the sphere n On the other side div F 3 yj z k and S is the sphere p a xi yj zk thus F n a and a JL 4 3 dV 3 7ra3 thus the two integrals are equal IJ 3 L n By V10 T H E DIVERGENCE THEOREM 3 Example 2 Use the divergence theorem to evaluate the flux of F x3 i across the sphere p a Solution Here div F 3 x2 y2 y3 j z3 k z2 3p2 Therefore by 2 we did the triple integration by dividing up the sphere into thin concentric spheres having volume dV 47rp2 dp Example 3 Let S1 be that portion of the surface of the paraboloid z 1 x2 y2 lying above the xy plane and let S2be the part of the xy plane lying inside the unit circle directed so the normal n points upwards Take F yz i xz j xy k evaluate the flux of F across S1by using the divergence theorem to relate it to the flux across S2 Solution We see immediately that div F 0 Therefore if we let Si be the same surface as S 2 but oppositely oriented so n points downwards the surface S1 Sh is a closed surface with n pointing outwards everywhere Hence by the divergence theorem Therefore since we have n k on S2 SIT by integrating in polar coordinates or by symmetry 2 Proof of the divergence theorem We give an argument assuming first that the vector field F has only a k component F P x y z k The theorem then says The closed surface S projects into a region R in the xy plane We assume S is vertically simple i e that each vertical line over the interior of R intersects S just twice S can have vertical sides however a cylinder would be an example S is then described by two equations 5 z g x y lower surface z h x y upper surface The strategy of the proof of 4 will be to reduce each side of 4 to a double integral over R the two double integrals will then turn out to be the same We do this first for the triple integral on the right of 4 Evaluating it by iteration we get as the first step in the iteration in 3 2 4 V VECTOR INTEGRAL CALCLUS To calculate the …
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