Question

Question: http://dl.dropbox.com/u/2792692/Question.png
So far, I have: r^3 dz dr dθ
Integrals are as followed:
(-pi/2) to (pi/2) for dθ
0 to 1 for dr
0 to [r^2*cos^2(θ)] for dz
Someone else told me that it is (r * cosθ) for dz

Answer 1

And that's great, all the questions I've answered from before are erased and I'm on level one (1)... Really?

Answer 2

The z term would be the same so let it as is
r would go from 0 to 1
\[\theta\] would go from 0 to 2pi

don't forget to include a r in integrand

Answer 3

x^2 + y^2 = r^2

with r
r^3

Answer 4

No, imran, x is always positive, so it's only in quadrants I and IV. -pi/2 to pi/2 for θ
Please, set it up before you answer.
Anyone know if the z thing is right?

Answer 5

sorry , you are right theta goes form -pi/2 to pi/2

Answer 6

let me type that up

Answer 7

Oh, I see why it's r cosθ! Haha, I wrote it down wrong on my paper as 0 to x^2 (for the original problem). Yeah, you're going to have to be better on these. I miss the old days when people actually cared about getting answers done on this site.

Answer 8

No, no, I don't need anymore help.

Answer 9

sorry , I don't provide answer if that's what you are looking for

Answer 10

Sorry, it looked as though you did (or tried to). Your comment didn't really do much but make me think if I was more wrong. So, it was not really help, but just random chatter.

Answer 11

You will go far with this attitude

Answer 12

You will go far with your math skills.

Answer 13

I definitely will