Recent content by uzman1243

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.

00 If possible, can you please explain what you mean from the graph that was plotted in post #7?

so the answer is just 72 pi?

r goes from 0 to 2 theta goes from 0 to pi/2 phi goes from 0 to pi correct?

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).

where are you getting the 144 from?

Sorry I made some changes a few minutes ago. Can you check my reply above again?

I cannot understand what you mean. Can you please explain a little more?

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...

ahhh so θ goes from 0 to pi ∅ goes from 0 to 2pi and r goes from 0 to 2 correct?

[FONT=Arial] So my method is valid? so ##dS## = N dr d ∅

ok so ∅ goes from 0 to 2pi as it is some sort of sphere/ elipse. correct?

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 ∫ ∫ ∫...