## anonymous one year ago ques

1. anonymous

|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

@IrishBoy123

3. anonymous

|dw:1441600974086:dw|

4. anonymous

|dw:1441601013362:dw|

5. anonymous

Nevermind I got it

6. anonymous

@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

I thought about divergence theorem, but then I've learnt about cylindrical coordinates yet

8. anonymous

I haven't*

9. anonymous

Oh!! I thought there were only 3 surfaces

10. anonymous

The answer is 90 so I thought S3 must be 0 for my result to make sense

11. anonymous

z=5 not 15 what I did was $-3y^2z=-15y^2z$

12. anonymous

sorry $-3y^2z=-15y^2$

13. IrishBoy123

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

That's something I tried earlier, but the answer is given as 90

15. anonymous

Your attempt looks absolutely perfect, I can't help but think now the that the answer is $-240\pi$

16. IrishBoy123

90 is through the curved surface....! i'll latex it in a moment.

17. anonymous

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

|dw:1441613005975:dw|

19. anonymous

"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

$0<z<5?$

21. IrishBoy123

$$= -5 \times 48 \pi = -240 \pi$$ 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

it says between z=0 and z=5, what does that imply to you?

23. IrishBoy123

nothing really hangs on that the answer is -240 pi or 90 depending on how you construe the question.

24. IrishBoy123

seems clear to me now that it is only the curved surface

25. anonymous

mmhm!!

26. anonymous

|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

got it!!!!

28. IrishBoy123

i can explain!!!

29. anonymous

$\vec F . \hat n=\frac{-16y-6y}{\sqrt{64+4y^2}}=\frac{-22y}{\sqrt{64+4y^2}}=\frac{-11}{\sqrt{16+y^2}}$ $ds=\frac{dydz}{|\hat i . \hat n|}=\frac{2\sqrt{16+y^2}}{8}dydz=\frac{\sqrt{16+y^2}}{4}dydz$ $\iint_\limits R \frac{-11}{4}dydz=\frac{-11}{4}\int\limits_{0}^{6} \int\limits_{0}^{4} dydz$

30. IrishBoy123

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

Oh!!!!!!! look at 2nd one :P

32. anonymous

I considered projection in yz plane

33. anonymous

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

35. IrishBoy123

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

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

is this a new one?

38. anonymous

yep! I'm getting perfect answer just a minus sign extra :(

39. IrishBoy123

make sure you normal points the right way!

40. IrishBoy123

|dw:1441615893452:dw|

41. IrishBoy123

|dw:1441615917894:dw|

42. anonymous

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

$\vec n = \pm <-8, 2y, 0>$ 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

here div F = 0

45. anonymous

$\vec \nabla \times \vec F=2\frac{\partial y}{\partial x}+0+0=2 \times \frac{\sqrt{8}}{2\sqrt{x}}=\sqrt{\frac{8}{x}}$ A positive x and negative y looks like an inward normal, don't we need an outward normal?

46. IrishBoy123

$<\partial_x, \partial_y, \partial_z>.<2y,-3,x^2> = 0$

47. anonymous

sorry I mean$\vec \nabla . \vec F$ not cross

48. IrishBoy123

$\partial_x (2y) + \partial_y(-3) + \partial_z(x^2) = 0$

49. anonymous

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

|dw:1441616713631:dw|

51. anonymous

|dw:1441616802795:dw|

52. IrishBoy123

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

are those the normals you're taking about?(look at the figure)

54. IrishBoy123

|dw:1441617005234:dw| i agree plug some numbers in to see

55. anonymous

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

|dw:1441617067457:dw|

57. IrishBoy123

so y is -ve, x =ve in the normal

58. IrishBoy123

|dw:1441617138630:dw|

59. IrishBoy123

gtg! good stuff!

60. anonymous

ok no biggie thanks a lot!

61. IrishBoy123

|dw:1441624438380:dw| pretty sure this is right