Calculating Total Charge on a Disk with Varying Charge Density

[Punkyc7]

Electric charge is distributed over a disk x^2+Y^2<=4 so that the charge density at (x,y) is

o(x,y)= x+y+x^2+y^2

what is the total charge on the disk

so I change to polar and get (rcos +rsin +r^2)rdrdtheta
and my limits go from 0 to 2pi and 0 to 2

for my answer i got 8 pi I am just wondering if that is right

 

[zhermes]

What's the question?

 

[Char. Limit]

He wants to know if he's correct in his answer.

 

[SammyS]

He wants to know if he's correct in his answer.

zhermes wants to know what was the question that Punkyc7 was answering when Punkyc7 asked if the answer was correct.

That's a valid inquiry by zhermes.

 

[HallsofIvy]

I would guess that Punkyc7 was asking if $8\pi$ is the total charge on the disc- though he did not specfically say that.

Punkyc7, I do NOT get $8\pi$. Show us what you did and we will decide which of us made a mistake.

 

[Punkyc7]

my question was it it correct, for the total charge on the disk


any ways

i integrated with respect to r and gotr^4/4+(r^3/3)cos+(+(r^3/3)sin evaluated from 0 to2



so i got 4+8/3 cos +8/3 sin and then i evaluted with respect to theta and got

8pi +8/3 sin-8/3 cos evaluated from 0 to 2pi


sin drops out and -8/3 -(8/3) drops out

so 8 pi that how i got my answer, i guess my question now is where did i go wrong

 

[Punkyc7]

is it in my limits or in my conversion to polar

 

[SammyS]

The cos(θ) term also drops out.

cos(2π) = cos(0)

 

[Dick]

I don't think there's any error. I get 8pi. The sin and cos terms both integrate to zero. So you've got just r^3*dr*dtheta. Looks like 8pi to me.

 

[zhermes]

I don't think there's any error. I get 8pi. The sin and cos terms both integrate to zero. So you've got just r^3*dr*dtheta. Looks like 8pi to me.

Absolutely correct. To elaborate a little... Punkyc7: if you look at the part of the equation, o(x,y)= x, this will be positive on one half of the disk, and just as much negative on the opposite side of the disk. Therefore the total is zero. The same argument applies to o(x,y) = y. Thus, as SammyS and dick point out, both of those terms fall out of the integral.

He wants to know if he's correct in his answer.

That's outrageously helpful, thanks.

 

[Char. Limit]

That's outrageously helpful, thanks.

Sorry that my attempt at answering your question fell so short of what you wanted. I'll be sure to try so much harder in the future.