Just need a check Finding total charge when given distribution

[xxbigelxx]

Homework Statement

We have a sphere (radius R) where charge is placed inside of it, ρ(r)=ρ0 cos(θ/3) sin(φ/2). Find the total charge. Your answer will be in terms of R and p0.

Homework Equations

The Attempt at a Solution

I think I did it correctly, but I would just like to make sure I did (and thus ensure my understanding). My answer is marked by the star, and everything below it is just my integration work that was required by the problem.

 

[gabbagabbahey]

Homework Statement

We have a sphere (radius R) where charge is placed inside of it, ρ(r)=ρ0 cos(θ/3) sin(φ/2). Find the total charge. Your answer will be in terms of R and p0.

Homework Equations

The Attempt at a Solution

I think I did it correctly, but I would just like to make sure I did (and thus ensure my understanding). My answer is marked by the star, and everything below it is just my integration work that was required by the problem.

I think you'll want to double check your integration of

\int_0^{\pi}\cos\left(\frac{\theta}{3}\right)\sin\theta d\theta

 

[xxbigelxx]

Hmm I tried to go over it again and I got the same result. Was it somewhere on the first integration by parts where I messed up? Thanks

 

[xxbigelxx]

Ohhh I was double checking the wrong part. I ended up making a mistake at the very bottom. It changed my result from that integration to 27/16 and my final answer is now...

(Rho)(R^3)(9/4) C

I hope this should be correct now...

 

[gabbagabbahey]

Ohhh I was double checking the wrong part. I ended up making a mistake at the very bottom. It changed my result from that integration to 27/16 and my final answer is now...

(Rho)(R^3)(9/4) C

I hope this should be correct now...

Looks good to me.

For future reference, you could have avoided IBP by using a trig identity:

\sin\theta\cos\left(\frac{\theta}{3}\right)=\frac{\sin\left(\frac{4\theta}{3}\right)+\sin\left(\frac{2\theta}{3}\right)}{2}

 

[xxbigelxx]

Ohh ok thanks. I don't even think I have ever come across that identity before.

 

[gabbagabbahey]

Ohh ok thanks. I don't even think I have ever come across that identity before.

Its just an application of the identity

\sin\theta\cos\left(\phi\right)=\frac{ \sin\left(\theta+\phi)+\sin\left(\theta-\phi\right)}{2}

 

[xxbigelxx]

Oh ok thanks a lot.