Nonuniformly charged spherical surface

A sphere of radius a in free space is nonuniformly charged over its surface such that the charge density is given by ρ_s(θ) = ρ_s0 sin 2θ, where ρ_s0 is a constant and 0≤θ≤∏. Compute the total charge of the sphere.

So I know
ρ_s = dQ/dS

Integrating the surface charge density function will give me the charge Q. My question is how would you set up this integral?

∫ρ_s0 sin 2θ dS
integrating 0 to ∏

Or would this involve much more than that such as a triple integral?

Any help getting this set up would be great! Thanks!

 

Alright then given the information above how would you set up the surface integral?

 

I haven't had calculus in a long time so I have forgotten most of it.

∫ ∫ ρs0 sin 2θ dA = ps0 sin 2θ * 8∏r dr dθ
0 -> ∏ 0 ->a

A = 4∏r²
dA = 8∏r

 

Well my first guess was a spherical coordinate system. Since it's a double integral it must be polar coordinates?

What would the limits then be?

 

Sorry I think you misunderstood me because that wasn't worded clearly. I originally thought that spherical coordinates were going to be used (before I posted this problem, using triple integrals) but since you stated it can be done using a double integral my thought was to use polar coordinates. You are given the limits for θ. Typically in polar you would use r,θ? So then all I would need to do is find the limits for r?

 

You are right and trying to pull it all back together is the hardest part. Does that involve parameterization described in terms of θ and Phi?

 

So then Phi from 0 to 2∏ and θ from 0 to ∏? Is this what you are saying?

 

Since r is constant then can I just say dS = r² sin θ dθ dPhi?

 

Alright that is good to know. So then should this be the integral to evaluate?

ρ_s0 * r²∫∫ sin(2θ) * sin(θ) dθ dPhi

0≤θ≤∏
0≤Phi≤2∏

After integrating that I get 2∏*ρ_s0*r².

 

For sin(2θ) * sin (θ) with respect to θ I got (2/3) sin(θ)^3.

Evaluated from 0 to ∏ I got 0.

Edit: TI89 got that answer which I have found isn't always reliable.