Question

I have another annoying question. Don't feel obligated to answer it

Answer 1

must answer question!

Answer 2

lol nooo the dont feel obligate to answer the question refers to u

Answer 3

nah, I'll let this one go, I don't have enough time...
it'll bother me though. you're doing this to me on purpose :P

Answer 4

If I knew anything to do with this particular portion of math, I'd be of assistance.

Answer 5

hmm u wanna lend me ur car for the day?

Answer 6

do you know divergence theorem? if so then we'll just have to evaluate:
\[\int\limits_{}^{} \int\limits_{}^{} \int\limits_{}^{}di v FdV\]

Answer 7

no i dont

Answer 8

havent learned it yet

Answer 9

aww man... The work could have been very very simple :)) all you had to do was to evaluate this:
\[ \int\limits_{0}^{2pi}\int\limits_{0}^{1}\int\limits_{1-r(\cos \theta+\sin \theta)}^{2-r(\cos \theta + \sin \theta)}3rdzdrd \theta\]

Answer 10

hahah well really all i neeed is to figure out the noraml vector and then the rest i can figure out on my own
Like its integral(F*NdA) so i just want to know how u firgure out the N

Answer 11

http://openstudy.com/users/samjordon#/updates/4f9b0637e4b000ae9ed083fd
umm like i have the solution here but i dont get how he got the N vector

Answer 12

ok so here, we have 3 parts of the surfaces that we shall evaluate and let's name them S1,S2,S3...
first is the part of the plane on the top which is x+y+z=2S1):
now remember that in writing an equation for a plane, the coefficients of x, y, and z will be the 1st, 2nd and third component of the vector normal to the plane. so the vector normal to the pplane is <1,1,1>. then turn this tto a unit vector so we'll have <1/sqrt(3),1/sqrt(3),1/sqrt(3)>. this will point away from the z axis so we'll leave it as is
S2: The cylinder part
ok this will need a little bit of intuition.....
notice that if I let a vector have no 3rd component and that the 1st and the 2nd components are x and y themselves respectively. Or in other words, at point (x1,y1,z1), <x1,y1,0> will be normal to the cylinder at that point. so a vector normal to the cylinder at any given point on the cylinder will be <x,y,0>. now we only need to turn this to a unit normal vector so just divide the components of the vector by the magnitude of the same vector so we'll arrive at <x/sqrt(x^2+y^2),y/sqrt(x^2+y^2),0> this will also point away from z axis so this we don't do nothing with.
S3: The part of x+y+z=1 below cylinder:
Doing the same steps and deduction we did with S1, the normal vector will be <1/sqrt(3),1/sqrt(3),1/sqrt(3)>. However, we need it to go away or point away from z axis so we'll just multiply this with -1 to have <-1/sqrt(3),-/sqrt(3),-1/sqrt(3)>

Answer 13

Wow :D K thanks for helping me out. This is great. U guys r amazing like idk how u have the patience to help me out