Can a Surface integral equal zero???

[Haris]

Hi!
It's a question from "Advanced Engineering Mathematics" by "Erwin Kreyszig" that is troubling me... Actually I'm getting the answer zero in evaluating following surface integral and I'm not sure whether I'm doing it right or wrong...

Q: Evaluate ∬(F.n dA) where
F=(x-z)i+(y-x)j+(z-y)k; S: r=[u*Cos(v) , u*Sin(v) , u] ; 0≤ u ≤5

Solution:
Since the surface is a cone, the interval of "v" would be 0≤ v ≤2(pi)

For n:
r_u=[Cos(v) , Sin(v) , 1]
r_v=[-u*Sin(v) , u*Cos(v) , 0]
r_u x r_v = n = [-u*cos(v) , -u*sin(v) , u]

F(r)=(u*cos(v)-u)i+(u*sin(v)-u*cos(v))j+(u-u*sin(v))k

∬(F.n dA) = ∬(-u²*cos²(v)+u²*cos(v)-u²*sin²(v)+u²*sin(v)*cos(v)+u²-u²*sin(v))dv*du limits: [0≤ u ≤5 ; 0≤ v ≤2(pi)]


-u²*cos²(v)-u²*sin²(v)+u²=0

therefore,

∬(F.n dA) = ∬(u²*cos(v)+u²*sin(v)*cos(v)-u²*sin(v))dv*du

Now, if I integrate with respect to "v" having limits zero to 2(pi) then the answer is zero...

I have re-checked it many times and still I end up with zero. Please check and point out the mistakes if any.

Thanx

 

[JohnZ]

you can use Divergence Theorem to check your result.

 

[JohnZ]

using Divergence (Gauss) Theorem, I get the result: 3 * V, V is volume of cone.

 

[Haris]

Thanx for your help JohnZ

According to your last thread, the result should be:

=3*V

Since volume of cone is (pi*r²*h)/3
=3*(pi*r²*h)/3

radius = 5; height =5;therefore

=392.7

This does not match my answer when evaluating the surface integral ... I checked it again and did not find any mistakes .... Please check out my first thread and explain why I am not getting the right answer.

Thanx

 

[JohnZ]

surface S is closed? if not, I should subtract 3*V by the surface integral on the cone's top, the result turns out to be zero.
In your thread, dA should be equal to u*du*dv in the polar coordinate system. But this won't change the result in your case.