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  1. anonymous
    • one year ago
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    |dw:1441598837930:dw| I have to find \[\iint_\limits S \vec A.\hat n ds\] Where \[\vec A=z \hat i+x \hat j-3y^2z \hat k\] For this I've divided the surface into 3 regions, s1 s2 and s3 |dw:1441599180575:dw| So we have \[\iint_\limits S \vec A . \hat n ds =\iint_{S_{1}} \vec A . \hat n ds+ \iint_{S_{2}}\vec A. \hat n ds+\iint_{S_{3}}\vec A. \hat nds\] For S1 we have \[\hat n=-\hat k\](outward normal) \[z=0\] and \[ds=dxdy\] \[\vec A . (-\hat k)=3y^2z\] So \[\iint_{S_{1}}\vec A . \hat n ds=\iint_{S_{1}}3y^2zdxdy=0\] because z is 0 in the xy plane For S3 we have \[\hat n=\hat k\] \[z=5\] \[ds=dxdy\] \[\vec A . \hat k=-3y^2z=-15y^2\] \[\iint_{S_{3}} \vec A . \hat n ds=\iint_{S_{3}} -15y^2dxdy=\int\limits_{0}^{4}\int\limits_{0}^{\sqrt{16-x^2}}-15y^2dydx=-5\int\limits_{0}^{4}(16-x^2)^\frac{3}{2}dx\] \[\iint_{S_{3}}\vec A.\hat n ds=-240\pi\] For S2 I considered it's projection on xz plane \[\iint_{S_{2}}\vec A .\hat n ds=\iint_\limits R \vec A.\hat n \frac{dxdz}{|(-\hat j).\hat n|}\] Here \[\hat n=\frac{x \hat i+y \hat j}{\sqrt{x^2+y^2}}=\frac{x \hat i+y \hat j}{4}\] so we have \[\vec A. \hat n=\frac{xz}{4}+\frac{xy}{4}=\frac{xz}{4}\] (y is 0 in xz plane) and \[ds=\frac{dxdy}{|(-\hat j).\hat n|}=\frac{\sqrt{x^2+y^2}}{y}dxdz=\frac{4}{y}dxdz\] \[\iint_\limits R \vec A . \hat n ds=\int\limits_{0}^{5}\int\limits_{0}^{4}\frac{xz}{y}dxdz\] Now since y becomes 0 this is kind of undefined and I'm stuck here :S

  2. anonymous
    • one year ago
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    @IrishBoy123

  3. anonymous
    • one year ago
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    |dw:1441600974086:dw|

  4. anonymous
    • one year ago
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    |dw:1441601013362:dw|

  5. anonymous
    • one year ago
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    Nevermind I got it

  6. anonymous
    • one year ago
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    @IrishBoy123 I found my mistake was suppose to put equation of y, I got surface integral through S2 as having a value of 90, but why isn't my surface integral through S3 equal to 0??

  7. anonymous
    • one year ago
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    I thought about divergence theorem, but then I've learnt about cylindrical coordinates yet

  8. anonymous
    • one year ago
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    I haven't*

  9. anonymous
    • one year ago
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    Oh!! I thought there were only 3 surfaces

  10. anonymous
    • one year ago
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    The answer is 90 so I thought S3 must be 0 for my result to make sense

  11. anonymous
    • one year ago
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    z=5 not 15 what I did was \[-3y^2z=-15y^2z\]

  12. anonymous
    • one year ago
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    sorry \[-3y^2z=-15y^2\]

  13. IrishBoy123
    • one year ago
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    i'll try divergence \[div \vec A = 0 + 0 -3y^2\] \[\iiint div \ \vec A \ dV\] \[=\int_{0}^{5} \ \int_{0}^{4} \ \int _{0}^{\sqrt{16-x^2}} -3y^2 \ dy \ dx \ dz\] \[= -\int_{0}^{5} \ \int_{0}^{4} \ \left[ y^3 \right]_{0}^{\sqrt{16-x^2}} \ dx \ dz\] \[= -\int_{0}^{5} \ \int_{0}^{4} \ (16-x^2)^{\frac{3}{2}} \ dx \ dz\] |dw:1441611982129:dw| \[= -5 \times 48 \pi\] not the answer you are suggesting.....

  14. anonymous
    • one year ago
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    That's something I tried earlier, but the answer is given as 90

  15. anonymous
    • one year ago
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    Your attempt looks absolutely perfect, I can't help but think now the that the answer is \[-240\pi\]

  16. IrishBoy123
    • one year ago
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    90 is through the curved surface....! i'll latex it in a moment.

  17. anonymous
    • one year ago
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    Yep, 90 is through the curved surface, I considered it's projection on xz plane to calculate it, so we have bottom=0, top=-240pi and curved surface in between as 90

  18. IrishBoy123
    • one year ago
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    |dw:1441613005975:dw|

  19. anonymous
    • one year ago
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    "S is the surface of the cylinder \[x^2+y^2=16\] included in the first octant between z=0 and z=5" Does this imply we only need to consider surface integral through the middle curved section?

  20. anonymous
    • one year ago
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    \[0<z<5?\]

  21. IrishBoy123
    • one year ago
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    \(= -5 \times 48 \pi = -240 \pi\) is for the whole lot. from the divergence theorem. that question could be construed as only through the curved surface and the answer for that is 90 but it is badly written .....

  22. anonymous
    • one year ago
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    it says between z=0 and z=5, what does that imply to you?

  23. IrishBoy123
    • one year ago
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    nothing really hangs on that the answer is -240 pi or 90 depending on how you construe the question.

  24. IrishBoy123
    • one year ago
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    seems clear to me now that it is only the curved surface

  25. anonymous
    • one year ago
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    mmhm!!

  26. anonymous
    • one year ago
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    |dw:1441613507846:dw| How about this?? \[F=2y\hat i-3\hat j+x^2 \hat k\] parabolic cylinder \[y^2=8x\] 1st octant and between planes y=4 and z=6 I'm thinking of projecting in xy plane \[\hat n=\frac{-8\hat i+2y \hat j}{\sqrt{64+4y^2}}\]

  27. IrishBoy123
    • one year ago
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    got it!!!!

  28. IrishBoy123
    • one year ago
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    i can explain!!!

  29. anonymous
    • one year ago
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    \[\vec F . \hat n=\frac{-16y-6y}{\sqrt{64+4y^2}}=\frac{-22y}{\sqrt{64+4y^2}}=\frac{-11}{\sqrt{16+y^2}}\] \[ds=\frac{dydz}{|\hat i . \hat n|}=\frac{2\sqrt{16+y^2}}{8}dydz=\frac{\sqrt{16+y^2}}{4}dydz\] \[\iint_\limits R \frac{-11}{4}dydz=\frac{-11}{4}\int\limits_{0}^{6} \int\limits_{0}^{4} dydz\]

  30. IrishBoy123
    • one year ago
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    divergence gives us the total figure for the whole volume, ir the end caps, the 2 rectangles and the flat surface. we got \(-240 \pi \) for that for S3 you also got \(-240 \pi \) which is correct. i messed my attempt up ;-( so good so far now we agree there's 90 through the curved surface if you look at the rectangles they amount to -40 and -50!!!!!! -40 in the xz plane, -50 in the zy plane IOW, i think we now own this problem :-))

  31. anonymous
    • one year ago
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    Oh!!!!!!! look at 2nd one :P

  32. anonymous
    • one year ago
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    I considered projection in yz plane

  33. anonymous
    • one year ago
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    I see now, -50 and -40 from the rectangles we get -90, and +90 from curved part this cancels out, bottom is 0 and from top we have -240pi, which is the total flux

  34. IrishBoy123
    • one year ago
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    exactly! i wish i had access to a 3D plotter we could see it happen

  35. IrishBoy123
    • one year ago
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    so you need to work out what they want from you, and it seems they just want the 90 through the curved surface

  36. anonymous
    • one year ago
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    Ooo I missed a y in my 2nd question lets see now \[\vec F . \hat n=\frac{-11y}{\sqrt{16+y^2}}\] so we have \[\vec F . \hat nds=-\frac{11y}{\sqrt{16+y^2}}.\frac{\sqrt{16+y^2}}{4}dydz=-\frac{11}{4}.ydydz\] \[-\frac{11}{4}\int\limits_{0}^{6}\int\limits_{0}^{4}ydydz=-\frac{11}{8}\int\limits_{0}^{6}[y^2]_{0}^{4}dz=-22\int\limits_{0}^{6}dz=-22 \times 6=-132\] but answer is 132 -_-

  37. IrishBoy123
    • one year ago
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    is this a new one?

  38. anonymous
    • one year ago
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    yep! I'm getting perfect answer just a minus sign extra :(

  39. IrishBoy123
    • one year ago
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    make sure you normal points the right way!

  40. IrishBoy123
    • one year ago
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    |dw:1441615893452:dw|

  41. IrishBoy123
    • one year ago
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    |dw:1441615917894:dw|

  42. anonymous
    • one year ago
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    from divergence theorem \[\iint_\limits S \vec F. \hat n ds=\iiint_\limits V \vec \nabla . \vec FdV=\sqrt{8}\int\limits_{0}^{6} \int\limits_{0}^{4} \int\limits_{0}^{\frac{y^2}{8}}x^{-\frac{1}{2}}dxdydz\] \[\implies 2\frac{\sqrt{8}}{\sqrt{8}}\int\limits_{0}^{6}\int\limits_{0}^{4}ydydz=16 \times 6=96\] This seems wrong O.o I don't see the point of taking our normal as i hat or -i hat since we will be taking mod?

  43. IrishBoy123
    • one year ago
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    \[\vec n = \pm <-8, 2y, 0>\] you want it with +ve x and -ve y so you \[\ <8, -2y, 0>\] it points away from the volume the || is only in the \(\frac{dx \ dy}{|\hat n \bullet \hat z|}\) but not in the \(\vec F \bullet \hat z\) div will give you the total for all surfaces in that volume, not just the one surface

  44. IrishBoy123
    • one year ago
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    here div F = 0

  45. anonymous
    • one year ago
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    \[\vec \nabla \times \vec F=2\frac{\partial y}{\partial x}+0+0=2 \times \frac{\sqrt{8}}{2\sqrt{x}}=\sqrt{\frac{8}{x}}\] A positive x and negative y looks like an inward normal, don't we need an outward normal?

  46. IrishBoy123
    • one year ago
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    \[ <\partial_x, \partial_y, \partial_z>.<2y,-3,x^2> = 0\]

  47. anonymous
    • one year ago
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    sorry I mean\[\vec \nabla . \vec F \] not cross

  48. IrishBoy123
    • one year ago
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    \[\partial_x (2y) + \partial_y(-3) + \partial_z(x^2) = 0\]

  49. anonymous
    • one year ago
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    but y is a function of x \[\frac{\partial y}{\partial x}=\frac{dy}{dx}=\frac{d(\sqrt{8x})}{dx}=\frac{\sqrt{8}}{2\sqrt{x}}=\frac{1}{2}\sqrt{\frac{8}{x}}\]

  50. IrishBoy123
    • one year ago
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    |dw:1441616713631:dw|

  51. anonymous
    • one year ago
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    |dw:1441616802795:dw|

  52. IrishBoy123
    • one year ago
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    you take the divergence of the field first then you relate that to the volume. only then dies the fact of the boundary matter

  53. anonymous
    • one year ago
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    are those the normals you're taking about?(look at the figure)

  54. IrishBoy123
    • one year ago
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    |dw:1441617005234:dw| i agree plug some numbers in to see

  55. anonymous
    • one year ago
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    That looks like inward to me, that's why I took the other normal, how would I know which is inward or outward?

  56. IrishBoy123
    • one year ago
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    |dw:1441617067457:dw|

  57. IrishBoy123
    • one year ago
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    so y is -ve, x =ve in the normal

  58. IrishBoy123
    • one year ago
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    |dw:1441617138630:dw|

  59. IrishBoy123
    • one year ago
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    gtg! good stuff!

  60. anonymous
    • one year ago
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    ok no biggie thanks a lot!

  61. IrishBoy123
    • one year ago
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    |dw:1441624438380:dw| pretty sure this is right

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