Compute the flux of the vector field,, through the surface, S.
and S is the surface
with
oriented in the positive y direction.
My answer:
I don't know if this is correct...
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Compute the flux of the vector field,
My answer:
I don't know if this is correct...
it looks good-however the orientatation seems a little strange.
you have a parabaloid aligned along the positive y axis
Usually we specify inward or outward--here on the half in the first and 8th octants you have an outward normal but in the 2d and 5th you have an inward normal.
So is this correct?
and of course your region of integration is a circle of radius 6 in the x-z plane
reason I ask is because I get;Quote:
Originally Posted by TheRekz
and I don't know how to represent this in polar
yes the ingrl is okay
in polar form just think of z as y
x= rcos(t)
z = r sin(t)
hmm.. so then it becomes:
am I missing something here?
I was using t for theta --just lazy
integral of cos(theta)^3 should not present a pblm
yes but i do need to multiply it by r??Quote:
Originally Posted by Calculus26
yes since x = rcos(theta) you'll end up integrating r^4 first
where does -432 come from?
because r is equal to 6, and therefore :Quote:
Originally Posted by Calculus26
x = rcos(theta)
= (6cos(theta))^3
= 216cos(theta)^3
multiplying that by -2, that's where I got the -432cos(theta)^3, am I doing something wrong here? so it should be -432cos(theta)^3r, right?
No r is a variable you take into account that r is 6 when you integrate r from 0 to 6
shouldn't we multiply it by r first and then take the integration with respect to r?
again x= rcos(theta) when you integrate over the disk of radius 6
r varies from 0 to 6 it isn't a constant 6
so you'd have (-2 r^4 cos^3(theta) +7) drd(theta) to integrate
well after that I got:
is this correct??
Yes
moving on forward I have:Quote:
Originally Posted by Calculus26
evaluated from 0 to 2pi.. resulting in 84pi and webassign doesn't take this answer