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I think you are integrating in this region. Not quite sure.
|dw:1389713649402:dw|

|dw:1389713913315:dw|

Obviously, \(0\le r\le 4\) and \(0\le \theta \le 2\pi\)

Mmm....0 <= r <= 2

Epic fail.... Yes Yes... Sorry :(

No worries, lol.

|dw:1389714352578:dw|

I am soooo dumb. I consider myself as quite new to this, although not so new...
Good luck!

Why does it look so complicated?! :S

What is the \(\phi_{\text{max}}\) you get?

Must you write in this.... ugly form?

Im sorry. Im just trying to answer this question......

What is the exact question?

Actually you need not write that ugly thing, I think. Let me double check something first.

alright

Do you agree that this is the region?
|dw:1389717244057:dw|

I figured it was the cone portion, yes.

(The cone part)

Ok, then no matter in which coordinate system, the cone is the region we want to integrate, right?

And \(\phi\) changes from 0 to \(\tan^{-1}(\frac{1}{2})\) as well.

Must you use that to integrate? It is really ugly and not necessary to do that step..

What if I can integrating it using spherical coordinates without getting to that step?

I mean to write that integral in spherical coordinates.

Yes.

How do you know you need to multiply \(\rho ^2 \sin\phi\) to the integral then?

Consider the black triangle
|dw:1389718930585:dw|

We get
\[\cos\phi = \frac{4}{r}\]\[r=\frac{4}{\cos\phi}=4 \sec\phi\]

Mhm, thats rho in the first integral, I managed that part, lol.

And \(\phi\) in the first part
|dw:1389719149953:dw|

Yes, arctan(1/2) because of the sides of the black triangle you drew.

I also have to go now. Will check this post again as soon as possible.

|dw:1389729705792:dw|