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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 xaxis, 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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This note was uploaded on 10/28/2010 for the course MA 261 taught by Professor Stefanov during the Fall '08 term at Purdue.
 Fall '08
 Stefanov
 Calculus

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