Flux Integrals. Solution. We want to visualize the surface together with the vector field. Here s a picture of exactly that:

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1 Flu Integrals The pictures for problems # - #4 are on the last page.. Let s orient each of the three pictured surfaces so that the light side is considered to be the positie side. Decide whether each of the following flu integrals is positie, negatie, or ero. ( F and G are the pictured ector fields.) (a) S F d S. Solution. We want to isualie the surface together with the ector field. Here s a picture of eactl that: (b) (c) As we can see, ectors in the ector field F that go through the surface S all go from the ellow side to the blue side. (We onl care about the ectors that actuall go through the surface; so, for instance, we can completel ignore the ectors in the bottom half of the picture since the don t go through the surface.) Since the surface is oriented so that the ellow side is considered to be the positie side, this means all of the ectors are going from the positie side to the negatie side, so the flu is negatie. F d S. Solution. Vectors in the ector field F that go through the surface go from the blue side to the ellow side. Since the surface is oriented so that the ellow side is considered to be the positie side, this means all of the ectors are going from the negatie side to the positie side, so the flu is positie. S 3 F d S. Solution. Vectors in the ector field F that go through the surface S 3 go from the blue side to the ellow side. Since the surface is oriented so that the ellow side is considered to be the positie side, this means all of the ectors are going from the negatie side to the positie side, so the flu is positie.

2 (d) G ds. S (e) (f) Solution. Vectors in the ector field G that go through the surface S go from the ellow side to the blue side. Since the surface is oriented so that the ellow side is considered to be the positie side, this means all of the ectors are going from the positie side to the negatie side, so the flu is negatie. G d S. Solution. None of the ectors in the ector field G actuall go through the surface, so the flu is ero. G ds. S 3 Solution. Vectors in the ector field G that go through the surface S 3 go from the ellow side to the blue side. Since the surface is oriented so that the ellow side is considered to be the positie side, this means all of the ectors are going from the positie side to the negatie side, so the flu is negatie. 2. In each part, ou are gien an orientation of one of the pictured surfaces. Decide whether this orientation means that the light side or dark side of the surface is the positie side, or if the description just doesn t make sense. (a) S, oriented with normals pointing upward. Solution. In order for the normals to be pointing upward, the must be sticking out of the ellow (light) side of the surface. (If the stuck out of the blue side, the would point downward.) So, the ellow (light) side is positie. (b), oriented with normals pointing upward. Solution. This doesn t make sense. If the normals at the top of the clinder point up, then the normals at the bottom must point down. There s no wa for all of the normal ectors to point upward. (c), oriented with normals pointing toward the -ais. Solution. This means that the ellow (light) side is positie. (d) S 3, oriented with normals pointing outward. Solution. This means that the blue (dark) side is positie. (e) S 3, oriented with normals pointing toward the origin. Solution. This means that the ellow (light) side is positie. 3. In each part, ou are gien a parameteriation of one of the three pictured surfaces. Decide whether the orientation induced b the parameteriation has the light side or dark side of the surface as the 2

3 positie side. (a) For S, r(u, ) = u,, u with u <. Solution. To figure out what the normals look like, we ll compute r u r : u r u =,, u2 + 2 r =,, u2 + 2 u r u r = u2 +, 2 u2 +, 2 Once we e computed r u r, it s not alwas eas to tell what the resulting orientation looks like. In this particular case, notice that the last component of r u r is just, which is of course alwas positie. This means that all of the normal ectors r u r point upwards. Looking at the picture of the surface S, this means that the ellow (light) side is positie. (Alternatiel, we could look back at #2(a), where we looked at this surface with upward normals.) (b) For S, r(u, ) = u cos, u sin, u with u < and < 2π. Solution. We first compute r u r : r u = cos, sin, r = u sin, u cos, r u r = u cos, u sin, u Notice that the last component of r u r is just u, and our parameteriation has u <. So, the normal ectors r u r point upwards. Looking at the picture of the surface S, this means that the ellow (light) side is positie. (c) For, r(u, ) = cos, u, sin with < u < and < 2π. Solution. We first compute r u r : r u =,, r = sin,, cos r u r = cos,, sin Here, it s not totall obious what s going on with the normal ectors. At the point r(u, ) = (cos, u, sin ), the normal ector points in the direction cos,, sin. One wa to figure out what s going on is just to find the normal ector at a particular point that s eas to isualie. For instance, let s tr to find the normal ector at the point (,, ). To do this, we need to first find the alues of u and corresponding to this point. That is, we hae to sole r(u, ) =,, for u and ; in this case, r(u, ) =,, is the same as saing cos =, u =, and sin =. Therefore, u = and = π 2. When u = and = π 2, the normal ector r u r is,,. This tells us that the normal ector at the point (,, ) on top of the clinder is just,,, which points upward. Looking at the picture of, this means that the blue (dark) side is positie. (d) For S 3, r(u, ) = sin cos u, sin sin u, cos with u < 2π and π. 3

4 Solution. We first compute r u r : r u = sin sin u, sin cos u, r = cos cos u, cos sin u, sin r u r = sin 2 cos u, sin 2 sin u, sin cos Since this looks a bit complicated, let s tr to just figure out the normal ector at a point we can understand easil, like the point (,, ). () To figure out the u and alues corresponding to this point, we sole r(u, ) =,, for u and. This gies three equations: sin cos u = sin sin u = cos = B the third equation (and the fact that π), we know that = π 2. Then, sin =, so the first two equations sa that cos u = and sin u =, which means u =. So, the point (,, ) corresponds to u =, = π 2. With these two alues, r u r =,,. That is, the normal ector at (,, ) points toward the origin. From our picture, this means that the ellow (light) side is positie. 4. Compute the following flu integrals (remember that parameteriations of the surfaces are gien in #3). Do the signs of our answers agree with our answers to #? (a) F ds, where S is oriented with normals pointing upward. ( F (,, ) =,,, as S before.) Solution. I like to organie m thinking into a few steps. Step - Parameterie the surface, and figure out the region in the u-plane describing the possible parameter alues. We were gien this information in #3; actuall, #3(a) and #3(b) gae us two different parameteriations. Let s use the parameteriation from #3(b), r(u, ) = u cos, u sin, u with u <, < 2π. (2) The region in the u-plane described b u <, < 2π is a rectangle: 2Π u () If ou tr the point (,, ) in this eample, ou ll find that the normal ector is,,. This happens occasionall, and it reall doesn t gie ou an useful information. Just pick another point and tr again. (2) Wh did I choose this one instead of the parameteriation from #3(a)? Well, in the parameteriation from #3(a), the region in the u-plane describing the possible parameters is a disk; in the parameteriation from #3(b), the region in the u-plane describing the possible parameters is a rectangle, which seems like an easier region of integration to deal with. 4

5 Step 2 - Decide whether the orientation gien b the parameteriation matches the desired orientation. In #3(b), we had alread figured out that this parameteriation had the ellow side as the positie side. The problem asks us to orient S with its normals pointing upward, which also means that the ellow side should be the positie side. So, the orientation described b the parameteriation does match the desired orientation. Step 3 - Compute! Since the orientation described b our parameteriation matches the orientation we want, we know the flu integral is F ds = F ( r(u, )) ( r u r ) da. S S To find F ( r(u, )), we just plug our parameteriation into F (,, ) =,,, which gies F ( r(u, )) =,, u. In #3(b), we found r u r = u cos, u sin, u. So, we hae F ds =,, u u cos, u sin, u da = u 2 da = 2π u 2 du d = 2π 3 Note that the sign matches our guess from #(a); we had said that, if the ellow side was considered the positie side (which it is, according to Step ), then the flu should be negatie. (b) G ds, where S2 is oriented with normals pointing toward the -ais. ( G(,, ) =,,, as before.) Solution. Let s follow the same three steps as in (a):. From #3(c), we hae the parameteriation r(u, ) = cos, u, sin with < u < and < 2π. The region in the u-plane described b these inequalities is a rectangle: 2Π u 5

6 2. In #3(c), we decided that this parameteriation had the blue side as the positie side, which does not match the described orientation (the described orientation has the ellow side as the positie side). 3. Since the orientation described b our parameteriation does not match the orientation we want, we know the flu integral is G ds = G( r(u, )) ( r u r ) da. We alread found in #3(c) that r u r = cos,, sin, so G ds =, u, cos,, sin da = da = Note that this matches our guess from #(e), where we guessed that the flu would be ero. (c) F ds, where S3 is oriented with normals pointing outward. ( F (,, ) =,,, as S 3 before.) Solution. Let s follow the same three steps as in (a):. From #3(d), we hae the parameteriation r(u, ) = sin cos u, sin sin u, cos with u < 2π and π. The region in the u-plane described b these inequalities is a rectangle: Π 2Π u 2. In #3(d), we decided that this parameteriation had the ellow side as the positie side, which does not match the described orientation (the described orientation has the blue side as the positie side). 3. Since the orientation described b our parameteriation does not match the orientation we want, we know the flu integral is F ds = F ( r(u, )) ( r u r ) da. S 3 6

7 S 3 We alread found in #3(d) that r u r = sin 2 cos u, sin 2 sin u, sin cos, so F ds =,, cos sin 2 cos u, sin 2 sin u, sin cos da = sin cos 2 da = = π 2π π = 2π 3 cos3 sin cos 2 du d 2π sin cos 2 d =π = = 4π 3 Note that the sign matches our answer from #(c): we had decided that, if the ellow side of the sphere was the positie side, then the flu would be positie. This means that, if the blue side of the sphere is positie (as it is in this problem), then the flu should be negatie. 5. Let S be the portion of the surface = 2 ling inside the clinder =, oriented with normals pointing upward. Let F (,, ) = 2,, 3 2. Ealuate the flu integral F ds. Solution. Let s follow the same three steps as in #4(a):. First, we need to parameterie the surface. If we rewrite the equation = 2 as = , then we see that we can parameterie it b r(u, ) = u,, 2 3u + 3. We want onl the portion with <. In terms of our parameters, this sas u <, so the region in the u-plane describing the possible parameter alues is the disk u < : S u 2. Net, we need to see what orientation this parameteriation describes, as we did in #3. First, we compute r u r : r u =,, 3 r =,, 3 r u r = 3, 3, This alwas points upward, so this matches the orientation we want (after all, we are told that the normals should point upward). 3. Now, we just compute. Since the orientation gien b r(u, ) matches the orientation we want, 7

8 we know the flu integral is F ds = F ( r(u, )) ( r u r ) da S = u 2,, 3 2 3, 3, da = ( 3u ) da Since is a disk, this integral will be easier to do in polar coordinates. In polar coordinates (thinking of u and as and ), the region is r <, θ < 2π, so the integral is 2π ( 2π 3r 2 r dr dθ = 3 ) r= 4 r4 dθ = 2π = 3 2 π 3 4 dθ r= 8

9 These are the surfaces for problems # - #4. Each is colored so that one side of the surface is light and the other side is dark. S is the portion of the cone = under the plane =. is the portion of the clinder = between the planes = and =. S 3 is the unit sphere =. These are the ector fields F and G for problems # - #4. (Note that the origin is located in the middle of each bo.) F (,, ) =,, G(,, ) =,, 9