The Divergence Theorem

Transcription

1 Math 1a The Divergence Theorem 1. Parameterize the boundary of each of the following with positive orientation. (a) The solid x + 4y + 9z 36. (b) The solid x + y z 9. (c) The solid consisting of all points (x, y, z) inside both the sphere x + y + z 4 and the cylinder x + y 3. Divergence (or Gauss ) Theorem: Let be a simple solid region and let be the surface of with positive orientation. Let F be a vector field with continuous partial derivatives. Then F d div F dv

2 . Let be the solid unit cube with opposing corners at the origin and (1, 1, 1) and faces parallel to the coordinate planes. Let be the boundary surface of, oriented with the outwardpointing normal. If F xy, 3ye z, x sin(z), find F d using the divergence theorem. (That is, find this flux integral by computing a triple integral instead.) x z y 3. Let F(x, y, z) x, y, e z. Let be the surface of the cube whose vertices are (±1, ±1, ±1), oriented with outward normals. valuate the flux integral F d.

3 4. Let F(x, y, z) x 3, z, 3y z. Let be the surface z x + y, z 4 together with the surface z 8 (x + y ), z 4. valuate the flux integral F d if is oriented with outward normals. 5. A friend tells you that if is a closed surface (that is, a surface without a boundary curve), then by tokes theorem we ought to have curl F d for any appropriate F (since there is no boundary curve C). Is this true? Can you justify it using the Divergence theorem? Hint: Remember that weird argument about ( F)?

4 6. Let be the solid bounded by the xy-plane and the paraboloid z 4 x y. Let be the boundary of (that is, piece of the paraboloid and a disk in the xy-plane), oriented with the outward-pointing normal. If F xz sin(yz) + x 3, cos(yz), 3zy e x +y, find F d using the divergence theorem. z y x 7. Here s a modification of the previous problem that adds a little clever trickery. Let s split up the surface into a union 1 of the piece 1 of the paraboloid and the flat disk, with both pieces oriented as in the previous problem. If the vector field F is the same as in the previous problem, find 1 F d, the flux of F through 1.

5 8. Let F be the vector field F(x, y, z) z 3 sin e y, z 3 e x sin z, y + z, and let be the bottom half of the sphere x + y + z 4, oriented with normals pointing upward. Find F d. (Hint: This is easier than it might seem at first) 9. Come up with some vector fields F(x, y, z) such that div F 1. Use this to compute the volume of a cone with height h and radius r.

6 1. Let be the part of the solid x +y ( z) with z 1. Calculate the volume of using the divervgence theorem. Notice that, the boundary of, typically needs to be broken into three pieces, so it would be ideal for F (r u r v ) to be simple (zero, for example) on one or two of these surfaces.

7 The Divergence Theorem Answers and olutions 1. We don t want to do the tedious work of parameterizing six surfaces (the six faces of the cube) in order to compute the flux integral directly. Instead we ll use the divergence theorem. Notice that div F ( ) xy + ) (3ye z + ( ) x sin(z) y + 3e z + x cos(z). x y z Thus the divergence theorem says that F d div F dv (y + 3e z + x cos(z)) dx dy dz 1 1 (y + 3e z + 1 ) cos(z) dy dz 1 (1 + 3e z + 1 ) cos(z) dz (e 1) + 1 sin(1). That looks like it would have been unpleasant to compute via six flux integrals.. This is even worse than the first problem. We couldn t possibly compute the flux integrals over the two pieces of the boundary (the paraboloid piece and the disk in the xy-plane) the vector field F is simply too complicated. Instead, we ll use the divergence theorem. Notice that div F ) (xz sin(yz) + x 3 + ( ) cos(yz) + (3zy e x +y ) x y z ) ( ) ) (z sin(yz) + 3x + z sin(yz) + (3y 3x + 3y. Thus we have from the divergence theorem F d div F dv D 4 x y ( 3x + 3y ) dz da, where D is the disk x + y 4 in the xy-plane. Thus we ll use polar coordinates for this double integral, or cylindrical coordinates for the triple integral: π 4 r ( F d ) 3r r dz dr dθ π π π ( 1r 3 3r 5) dr dθ [ 3r 4 1 ] r6 dθ (48 3) dθ 3π.

8 That wasn t bad at all. 3. The difficulty here is that 1 is not the boundary surface of a simple solid region, so we can t blindly apply the divergence theorem. The standard trick is to add another piece to the boundary so that 1 bounds a region. This was already done for us in the previous problem, and in fact we found that 3π F d F d F d. Thus to find the requested flux we can either integrate over 1 (icky!) or (much nicer). We parameterize with polar coordinates, so r(r, θ) r cos(θ), r sin(θ), (with θ π and r ). Then i j k r θ r r r sin(θ) r cos(θ),, r. cos(θ) sin(θ) Note that this vector points downward, with is the appropriate orientation for (and why we chose r θ r r rather than r θ r θ ). Thus the flux through is π F d F(r(r, θ)),, r dr dθ π π π e 4 1 r 3 cos 3 (θ), 1, e r,, r dr dθ re r dr dθ π dθ π e4 1 Thus the flux through is 3π π (e 4 1) π (33 e 4 ). e r dθ π ( e 4 1 ). 4. This is true. We can justify it with the divergence theorem by recalling that div(curl F). We proved this in the Curl and Divergence worksheet and tewart does it in that section as well. The point of the second hint is that u (u v), since u v is perpendicular to u, so you can remember that ( F) by analogy. In any case, if is a closed surface, then we can let be the solid enclosed by. Then the divergence theorem says that curl F d div (curl F) dv dv.

9 5. (a) If div F > at P, then a F d > as well, at least when a is small enough. Thus we would expect flux out of P. (b) If div F < at P, a similar argument to part (a) says that we would expect more flux into P. (c) When div F at P, we expect that a F d. This means that the same amount of flow in toward P as out from P. 6. (a) We re aiming to use the divergence theorem, so we compute the divergence of F: div F ( x 3y z ) + x y (exz y) + (z cos(xy)) z Thus the divergence theorem implies that F d div F dv dv Vol() 8 3 πa3. (b) What this problem is getting at is that if we have a vector field F with div F 1, then we can compute the volume of via the flux over the boundary : Vol() 1 dv div F dv F d. We ll compute this flux with the given F 1 x, y, z. 3 We parameterize the sphere of radius a with r(φ, θ) a sin(φ) cos(θ), a sin(φ) sin(θ), a cos(φ), from which we find that i j k r φ r θ a cos(φ) cos(θ) a cos(φ) sin(θ) a sin(φ) a sin(φ) sin(θ) a sin(φ) cos(θ) a sin(φ) sin(φ) cos(θ), sin(φ) sin(θ), cos(φ) a sin(φ) a sin(φ) cos(θ), a sin(φ) sin(θ), a cos(φ) a sin(φ) r(φ, θ) is the outward-pointing normal. ince F 1r and r a, we get F (r 3 φ r θ ) a sin(φ) r a3 sin(φ). Thus the flux is 3 3 π π a 3 sin(φ) π π a 3 4πa3 F d dφ dθ dθ As expected, this flux is the volume of.

10 7. As in Problem 6(b), we will compute the volume by computing the flux. This time (as suggested) we ll use F 1 x, y,, so div F 1. We split into three pieces, the top, bottom, and sides: 1, the top, the bottom 3, the sides Notice that is the union 1 3, so Vol() F d F d + F d F d. Two of these three new flux integrals are really simple. Notice that the normals for 1 and are either n k,, 1 or n k. Because of the clever way we chose F, however, we thus get F d! o the flux integrals over 1 and are zero. On 3 we need to actually compute the flux integrals. This involves parameterizing the surface, given to us as x + y ( z). Thus, for fixed z, this is a circle of radius z. This means we have the parameterization r(z, θ) ( z) cos(θ), ( z) sin(θ), z, and so i j k r θ r z ( z) sin(θ) ( z) cos(θ) cos(θ) sin(θ) 1 ( z) cos(θ), ( z) sin(θ), ( z) is the appropriately oriented normal. ince F(r(z, θ)) 1 ( z) cos(θ), ( z) sin(θ),, we get the flux integral 3 F d π 1 1 π [ ( z) dz dθ 1 ] 1 π 6 ( 7 z)3 dθ 6 dθ 7π 3. Finally, we have found the volume of : + + 7π 7π. (Notice that this agrees with the 3 3 usual formula for the volume of a cone: V 1 3 πr h. The full cone has r h, while the removed cone has r h 1, so the truncated cone has volume 1 3 π3 1 3 π13 7π.) 3