Question

A cone has its tip at the point (0,0,5) and its base the disk D, x^2+y^2<=4, in the xy-plane. The surface of the cone is the curved and slanted face, S, oriented upward, and the flat base, D, oriented downward. The flux of the constant vector field F=ai+bj+czk through S is given by integral(of S) of F dA =5.48. What is the integral(of D) of F dA equal to?

I'm stuck :(

Answer 1

Lol, any help on this is much appreciated. Calc III hw is killing me

Answer 2

hmm, I don't remmember exactly, but i thin that the flux of constant field over closed volume shouldn't be 0? That would give -5.48 for the D part

Answer 3

If F = <a, b, c> then the flux over just D would be -5.48. That was the first part of the question. The second part states that F now equals <a, b, cz> so it's not techincally constant, and I apologize for the wording.

Answer 4

ok... np. let me think, :)

Answer 5

The Field is not constant

Answer 6

z is perpendicular to D. So just simple integral in 1 variable

Answer 7

Is there a misprint?

Answer 8

myko....what would that integral be then in this case?

Answer 9

Is the field
F=ai+bj+czk
or
F=ai+bj+ck

Answer 10

eliasaab...F=<a, b, cz> as was stated in my reply a few moments ago. Sorry for the confusion.

Answer 11

Great, that makes it easy. Stay tuned.

Answer 12

What is the volume of your cone? DO you know hoe to do that?

Answer 13

Well the the tip of the cone is (0,0,8) and the base is in the form of x^2 + y^2 <=9, so V=(1/3)(pi)*(r^2)*h where r=3 and h=8?

Answer 14

\[
V=\frac{1}{3} \pi h r^2
\]

Answer 15

r =2 and h=5

Answer 16

Err, yeah. The numbers on the problem I posted are different from my actual problem on the homework. So yeah, lets just go with r=2, h=5

Answer 17

What's the divergence of your field?

Answer 18

divF = c am i right?

Answer 19

\[fluxD = \int\limits \int\limits _{S}zkdxdy\]
to solve this parametrize the circle

Answer 20

myko, where did u get the integrand "zk dx dy" ?

Answer 21

Yes
Flux of your field
\[
Flux= \int \int \int_{Cone} \text { Divergence (F) } dV

\]

Answer 22

I did try doing what you just did, eliassab, but that's just giving my the flux from the cone, which is in this case 5.48. I'm guessing solve for c from that integral?

Answer 23

That is what i got, yes..

Answer 24

Since you know c, it easy to compute the flux around the base.

Answer 25

Your filed is now known. Apply the definition of the flux and you are done

Answer 26

across the base

Answer 27

Flux at the base = c* pi(r^2)?

Answer 28

- c pi r^2 since n=(0,0,-1}

Answer 29

Oh okay, let me put this in to my web hw and see if it's all good.

Answer 30

Input the right numbers and not the numbers posted.

Answer 31

so I did try doing this, where in my case h = 8 and r = 3

Answer 32

solving for c, I got 3.35, and thus the flux at the base should've been -c*pi*3^2

Answer 33

giving me -94.719

Answer 34

but the answer is wrong...

Answer 35

Repost the the exact problem and let me see.

Answer 36

A cone has its tip at the point (0,0,8) and its base the disk D, x^2+y^2<=9, in the xy-plane. The surface of the cone is the curved and slanted face, S, oriented upward, and the flat base, D, oriented downward. The flux of the t vector field F=ai+bj+czk through S is given by integral(of S) of F dA =3.12. What is the integral(of D) of F dA equal to?

Answer 37

I got 3.3518

Answer 38

then according to the equation you gave me where (1/3)c(pi)(h) = r^2(flux given)

Answer 39

Lol, so c = .01379 about?

Answer 40

so c= .04138 ?

Answer 41

then -c(pi*r^2) gives the flux through D, where c=.04138?

Answer 42

yes

Answer 43

alright, let me try

Answer 44

it is about -1.17

Answer 45

Hmm, still saying it's wrong.

Answer 46

I know what went wrong. Stay tuned.

Answer 47

The 3.12 is the flux thru S and not thru the entire outside area of the cone. Hence
\[
3.12 -\pi c r^2=c \frac{1}{3}
\pi h r^2

\]
Solve for c, you get
\[
c\approx .03
\]
your answer is
\[
-c \pi r^2 =-0.85
\]