16.9 - Section 17.9 The Divergence Theorem “Turning a...

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Unformatted text preview: Section 17.9 The Divergence Theorem “Turning a Flux Integral into a Triple Integral” The last result we consider is a generalization of Green’s Theorem converting a flux integral over a closed surface into a triple integral over the interior of the surface. 1. The Divergence Theorem To state the divergence theorem, we need the following definition which uses the ideas we built up in the section on triple integrals. Definition 1.1. A solid E is called a simple solid region if it is one of the types (either Type 1, 2 or 3) given in Section 16.6. Examples of a simple solid regions are spheres, ellipsoids, portions of cylinders etc. We can now state the divergence theorem. Result 1.2. Let E be a simple solid region and let S be the boundary of E with outward orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region which contains E . Then F · dS = S E divF dV. This Theorem tells us that any flux integral over a closed surface can be converted into a triple integral over the interior (provided it is defined on the interior), and since flux integrals in general are much more difficult than triple integrals, this gives us an easier way to calculate them. We illustrate with some examples. Example 1.3. (i ) Use the divergence theorem to evaluate (ex sin (y )i + ex cos (y )j + yz 2 k ) · dS S where S is the box bounded by the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. Applying the divergence theorem, we have 1 1 0 1 0 1 1 (e sin (y )i+e cos (y )j +yz k )·dS = S 1 1 0 0 1 1 0 x x 2 (ex sin (y )−ex sin (y )+2yz )dxdydz zdz = 0 = 0 2yzdxdydz = 0 y 2z dz = 0 1 2 1 2 (ii ) Suppose F (x, y, z ) = x (x2 + y2 + z2) 2 3 i+ y (x2 + y2 + z2 ) 2 3 j+ z (x2 x + y2 + z2) 2 3 k and S is the cylinder y 2 + z 2 = 4 with −2 the integral F · dS S 2. Show that is the same as the integral F · dS S1 where S1 is the unit sphere centered at the origin oriented outward. Observe that divF = 0 (if you are unsure, check). However, we cannot apply the divergence theorem since F is not defined at (0, 0, 0). Observe however that if S1 is the unit sphere centered at the origin, and S2 is the closed surface which consists of S (oriented outward) and S1 (oriented inward), we can apply the divergence theorem to get F · dS = S2 I nt(S2 ) 0dV = 0. However, F · dS = S2 S F · dS − S1 F · dS = 0 F · dS. so F · dS = S S1 (iii ) Use the divergence theorem to evaluate (3xy 2 i + xez j + z 3 k ) · dS S where S is the surface of the cylinder y 2 + z 2 = 1 and the planes x = −1 and x = 2. We have divF = 3y 2 + 3z 2 , so applying the divergence theorem, we have (3xy 2 i + xez j + z 3 k ) · dS = S R (3y 2 + 3z 2 )dV. Converting to cylindricals along the x-axis, we have 3y 2 +3z 2 = r 2 , 0 r 1, 0 ϑ 2π , and −1 x 2, so 2 2π 0 0 1 3y 2 + 3z 2 dV = R −1 3 9π 3r 3drdϑdx = 2π (3)( ) = 4 2 3 As with Green’s Theorem, and Stokes Theorem, there are ways to apply the divergence theorem indirectly. We illustrate with some examples. Example 1.4. Let S be the open cone z = Calculate F · dS S (x2 + y 2) with z 3. for each of the following: (i ) F = xi + y j + z k (ii ) F = xi + y j We consider each problem individually. (i ) Consider the surface L which consists of S and the cap C on top of the cone oriented upward. Then the divergence Theorem implies F · dS = L I 3dV where I is the interior of the cone. Therefore, we get 1 F · dS = 3dV = 3 × π 9 × 3 = 27π 3 L I i.e. 3 times the volume of the cone since the integral of the function 1 determines the volume of the region. This means F · dS = 27π − S C F · dS the latter integral which is much easier than the first. In particular, the cap is the part of the function g (x, y ) = 3 over the circle D given by x2 + y 2 9, so we can use the flux formula for graphs of functions. Specifically, we have ∂g ∂g F · dS = (−x −y + 3)dA = 3dA = 27π ∂x ∂y C D D i.e. 3 times the area inside the circle D . In particular, we have F · d S = 0. S Note that we could have simply concluded this same result by observing that the vectors in the vector field F point along the surface S and do not flow through the surface! (ii ) We can follow a similar process as in the previous example. In particular, in this case, we shall get F · dS = 18π − S C F · dS. Observe however that F · dS = 0 C 4 since all the vectors in the vector field F point along the surface C (and not through the surface C . Hence F · dS = 18π. S We have learnt a lot of new material over the last few days, and as with line integrals, we have learnt many different ways to calculate them. Therefore, we finish, as we did with line integrals, with a decision tree of which questions to ask when trying to determine which method to use. As before, the general idea is to avoid using direct calculation as much as possible, and instead convert the problem using either Stokes theorem or the divergence theorem into a much simpler problem. Notice in the decision tree that only one path leads to direct calculation (at all costs, we want to avoid direct calculation since it is always a difficult process!). ooo No oooo o ooo wooo i iiii No iiiii i iii iiii tiiii Is S closed? QQ QQQ QQQYes QQQ QQQ Q( Is F a Curl field? Yes Is F divergent free? V No   VVVV VVVV Yes VVVV VVVV VVVV V+ S Indirect Divergence or indirect Stokes No Stokes Theorem Divergence Theorem F dS = 0 by the Divergence  Direct Calculation 5 ...
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