A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
ques
anonymous
 one year ago
ques

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441598837930:dw I have to find \[\iint_\limits S \vec A.\hat n ds\] Where \[\vec A=z \hat i+x \hat j3y^2z \hat k\] 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 \[\hat n=\hat k\] \[z=5\] \[ds=dxdy\] \[\vec A . \hat k=3y^2z=15y^2\] \[\iint_{S_{3}} \vec A . \hat n ds=\iint_{S_{3}} 15y^2dxdy=\int\limits_{0}^{4}\int\limits_{0}^{\sqrt{16x^2}}15y^2dydx=5\int\limits_{0}^{4}(16x^2)^\frac{3}{2}dx\] \[\iint_{S_{3}}\vec A.\hat n ds=240\pi\] 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441600974086:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441601013362:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought about divergence theorem, but then I've learnt about cylindrical coordinates yet

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh!! I thought there were only 3 surfaces

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The answer is 90 so I thought S3 must be 0 for my result to make sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0z=5 not 15 what I did was \[3y^2z=15y^2z\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry \[3y^2z=15y^2\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i'll try divergence \[div \vec A = 0 + 0 3y^2\] \[\iiint div \ \vec A \ dV\] \[=\int_{0}^{5} \ \int_{0}^{4} \ \int _{0}^{\sqrt{16x^2}} 3y^2 \ dy \ dx \ dz\] \[= \int_{0}^{5} \ \int_{0}^{4} \ \left[ y^3 \right]_{0}^{\sqrt{16x^2}} \ dx \ dz\] \[= \int_{0}^{5} \ \int_{0}^{4} \ (16x^2)^{\frac{3}{2}} \ dx \ dz\] dw:1441611982129:dw \[= 5 \times 48 \pi\] not the answer you are suggesting.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's something I tried earlier, but the answer is given as 90

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your attempt looks absolutely perfect, I can't help but think now the that the answer is \[240\pi\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.190 is through the curved surface....! i'll latex it in a moment.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, 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

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441613005975:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"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?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(= 5 \times 48 \pi = 240 \pi\) 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 .....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it says between z=0 and z=5, what does that imply to you?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1nothing really hangs on that the answer is 240 pi or 90 depending on how you construe the question.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1seems clear to me now that it is only the curved surface

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441613507846:dw How about this?? \[F=2y\hat i3\hat j+x^2 \hat k\] parabolic cylinder \[y^2=8x\] 1st octant and between planes y=4 and z=6 I'm thinking of projecting in xy plane \[\hat n=\frac{8\hat i+2y \hat j}{\sqrt{64+4y^2}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\vec F . \hat n=\frac{16y6y}{\sqrt{64+4y^2}}=\frac{22y}{\sqrt{64+4y^2}}=\frac{11}{\sqrt{16+y^2}}\] \[ds=\frac{dydz}{\hat i . \hat n}=\frac{2\sqrt{16+y^2}}{8}dydz=\frac{\sqrt{16+y^2}}{4}dydz\] \[\iint_\limits R \frac{11}{4}dydz=\frac{11}{4}\int\limits_{0}^{6} \int\limits_{0}^{4} dydz\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1divergence 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 :))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh!!!!!!! look at 2nd one :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I considered projection in yz plane

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1exactly! i wish i had access to a 3D plotter we could see it happen

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so you need to work out what they want from you, and it seems they just want the 90 through the curved surface

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ooo 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 _

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1is this a new one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep! I'm getting perfect answer just a minus sign extra :(

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1make sure you normal points the right way!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441615893452:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441615917894:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from 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?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\[\vec n = \pm <8, 2y, 0>\] 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.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?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\[ <\partial_x, \partial_y, \partial_z>.<2y,3,x^2> = 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry I mean\[\vec \nabla . \vec F \] not cross

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\[\partial_x (2y) + \partial_y(3) + \partial_z(x^2) = 0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but y is a function of x \[\frac{\partial y}{\partial x}=\frac{dy}{dx}=\frac{d(\sqrt{8x})}{dx}=\frac{\sqrt{8}}{2\sqrt{x}}=\frac{1}{2}\sqrt{\frac{8}{x}}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441616713631:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441616802795:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you take the divergence of the field first then you relate that to the volume. only then dies the fact of the boundary matter

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are those the normals you're taking about?(look at the figure)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441617005234:dw i agree plug some numbers in to see

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That looks like inward to me, that's why I took the other normal, how would I know which is inward or outward?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441617067457:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so y is ve, x =ve in the normal

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441617138630:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok no biggie thanks a lot!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441624438380:dw pretty sure this is right
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.