Evaluate both sides of divergence theorem

[acsol]

Homework Statement

NOTE: don't know see the phi symbol so I used theta. this is cylindrical coordinates not spherical.

Given the field D = 6ρsin(θ/2)a_p + 1.5ρcos(θ/2)a_θ C/m^2 , evaluate both sides of the divergence theorem for the region bounded by ρ=2, θ=0 to ∏, and z = 0 to 5

Homework Equations

∫D * dS = ∫ ∇ * D dV

The Attempt at a Solution

The answer (to both sides of course) is 225. I figured out how to do the divergence side, but I'm having problems evaluating the left (surface) side. The way I'm doing it is that the z-component is zero since the equation for D only has a rho and phi.

From here I'm evaluating ∫∫D_p*pdθdZ from 0 to pi for dθ and 0 to 5 for dz.

I plug in p = 2 as an "initial condition", and when I calculate the integral through I keep getting 240 even though the answer is 225. Any ideas what's going on? Maybe my integral is just wrong

 

[Dick]

Homework Statement

NOTE: don't know see the phi symbol so I used theta. this is cylindrical coordinates not spherical.

Given the field D = 6ρsin(θ/2)a_p + 1.5ρcos(θ/2)a_θ C/m^2 , evaluate both sides of the divergence theorem for the region bounded by ρ=2, θ=0 to ∏, and z = 0 to 5

Homework Equations

∫D * dS = ∫ ∇ * D dV

The Attempt at a Solution

The answer (to both sides of course) is 225. I figured out how to do the divergence side, but I'm having problems evaluating the left (surface) side. The way I'm doing it is that the z-component is zero since the equation for D only has a rho and phi.

From here I'm evaluating ∫∫D_p*pdθdZ from 0 to pi for dθ and 0 to 5 for dz.

I plug in p = 2 as an "initial condition", and when I calculate the integral through I keep getting 240 even though the answer is 225. Any ideas what's going on? Maybe my integral is just wrong

You are only integrating over part of the surface. You have half a cylinder, it also has a top, a bottom and a flat side. You are only doing the rounded side. Can you picture it? What other part of the surface might contribute to your integral?

 

[acsol]

Hmm. I thought about that but my answer didn't really change and here's why:

The cylinder has 4 sides, right? Rounded, opposite of rounded (flat), top, and bottom. However, from the given equation for D, we only have a rho and phi component. That automatically makes the top and bottom zero.

WAIT, are you saying that I need to do a coordinate conversion and then calculate the surface dxdz?

 

[Dick]

Hmm. I thought about that but my answer didn't really change and here's why:

The cylinder has 4 sides, right? Rounded, opposite of rounded (flat), top, and bottom. However, from the given equation for D, we only have a rho and phi component. That automatically makes the top and bottom zero.

WAIT, are you saying that I need to do a coordinate conversion and then calculate the surface dxdz?

Exactly. That's where you'll find the -15 missing units of surface flux. Very sharp of you to figure out how to do it. You are almost there.

 

[acsol]

Took a while but got it right

Thanks for the help mate!