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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
|dw:1441600974086:dw|

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|dw:1441601013362:dw|
Nevermind I got it
@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??
I thought about divergence theorem, but then I've learnt about cylindrical coordinates yet
I haven't*
Oh!! I thought there were only 3 surfaces
The answer is 90 so I thought S3 must be 0 for my result to make sense
z=5 not 15 what I did was \[-3y^2z=-15y^2z\]
sorry \[-3y^2z=-15y^2\]
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.....
That's something I tried earlier, but the answer is given as 90
Your attempt looks absolutely perfect, I can't help but think now the that the answer is \[-240\pi\]
90 is through the curved surface....! i'll latex it in a moment.
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
|dw:1441613005975:dw|
"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?
\[0
\(= -5 \times 48 \pi = -240 \pi\) 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 .....
it says between z=0 and z=5, what does that imply to you?
nothing really hangs on that the answer is -240 pi or 90 depending on how you construe the question.
seems clear to me now that it is only the curved surface
mmhm!!
|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 \[\hat n=\frac{-8\hat i+2y \hat j}{\sqrt{64+4y^2}}\]
got it!!!!
i can explain!!!
\[\vec F . \hat n=\frac{-16y-6y}{\sqrt{64+4y^2}}=\frac{-22y}{\sqrt{64+4y^2}}=\frac{-11}{\sqrt{16+y^2}}\] \[ds=\frac{dydz}{|\hat i . \hat n|}=\frac{2\sqrt{16+y^2}}{8}dydz=\frac{\sqrt{16+y^2}}{4}dydz\] \[\iint_\limits R \frac{-11}{4}dydz=\frac{-11}{4}\int\limits_{0}^{6} \int\limits_{0}^{4} dydz\]
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 :-))
Oh!!!!!!! look at 2nd one :P
I considered projection in yz plane
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
exactly! i wish i had access to a 3D plotter we could see it happen
so you need to work out what they want from you, and it seems they just want the 90 through the curved surface
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 -_-
is this a new one?
yep! I'm getting perfect answer just a minus sign extra :(
make sure you normal points the right way!
|dw:1441615893452:dw|
|dw:1441615917894:dw|
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?
\[\vec n = \pm <-8, 2y, 0>\] 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
here div F = 0
\[\vec \nabla \times \vec F=2\frac{\partial y}{\partial x}+0+0=2 \times \frac{\sqrt{8}}{2\sqrt{x}}=\sqrt{\frac{8}{x}}\] A positive x and negative y looks like an inward normal, don't we need an outward normal?
\[ <\partial_x, \partial_y, \partial_z>.<2y,-3,x^2> = 0\]
sorry I mean\[\vec \nabla . \vec F \] not cross
\[\partial_x (2y) + \partial_y(-3) + \partial_z(x^2) = 0\]
but y is a function of x \[\frac{\partial y}{\partial x}=\frac{dy}{dx}=\frac{d(\sqrt{8x})}{dx}=\frac{\sqrt{8}}{2\sqrt{x}}=\frac{1}{2}\sqrt{\frac{8}{x}}\]
|dw:1441616713631:dw|
|dw:1441616802795:dw|
you take the divergence of the field first then you relate that to the volume. only then dies the fact of the boundary matter
are those the normals you're taking about?(look at the figure)
|dw:1441617005234:dw| i agree plug some numbers in to see
That looks like inward to me, that's why I took the other normal, how would I know which is inward or outward?
|dw:1441617067457:dw|
so y is -ve, x =ve in the normal
|dw:1441617138630:dw|
gtg! good stuff!
ok no biggie thanks a lot!
|dw:1441624438380:dw| pretty sure this is right

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