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|dw:1441600974086:dw|

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Nevermind I got it

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 \[-3y^2z=-15y^2\]

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.

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\[0

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!!

got it!!!!

i can explain!!!

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

I considered projection in yz plane

exactly!
i wish i had access to a 3D plotter
we could see it happen

is this a new one?

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

make sure you normal points the right way!

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here div F = 0

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

sorry I mean\[\vec \nabla . \vec F \]
not cross

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

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

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

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

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gtg!
good stuff!

ok no biggie thanks a lot!

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pretty sure this is right