Thank you so much. Finally makes sense.
Just one more question, how do you know if a surface is closed? So that I could have used the divergence theorem here instead.
I don't get it.
using divergence theorem I get 216pi. Using the surface integrals I get 72pi. The answer to solving using surface integrals should give me the same answer as using divergence theorem (provided it's a close surface).
Ok so I am trying the surface integral
r(θ,z) = 3cosθ i + 3sinθ j + zk
dr/dθ = -3sinθ i + 3cosθ j +0 k
dr/dz = 0i + 0j + 1k
dr/dθ x dr/dz = 3cosθ, 3sinθ,0
|dr/dθ x dr/dz| = 3
thus I get $$\vec n$$ = cosθ,sin θ,0
finally adding into the integral
∫∫ (3zcosθ,3cosθ,z2).(cosθ,sinθ,0) .3 dθ dz...
x goes from -2 to 2
y goes from 0 to √4-x2 circle with radius 2
z from 0 to √4-x2-y2 sphere with with radius 2
so I am guessing
r goes from 0 to 2
∅ and θ from 0 to 2π
Homework Statement
Homework EquationsThe Attempt at a Solution
I thought of using the divergence theorem where
I find that ∇.F = 3z
thus integral is
∫ ∫ ∫ 3z r dz dr dθ where r dz dr dθ is the cylindrical coordinates
with limits
0<=z<=4
0<=r<=3
0<=θ<=2π
and solving gives me 216π
Can I...
Homework Statement
The question asks me to convert the following integral to spherical coordinates and to solve it
Homework EquationsThe Attempt at a Solution
just the notations θ = theta and ∅= phi
dx dy dz = r2 sinθ dr dθ d∅
r2 sinθ being the jacobian
and eventually solving gets me
∫ ∫ ∫...