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|dw:1352751755636:dw|

1st octant...

Make the equations equal to one another...

yes, and what do you get for theta?

arccos 2^(1/2)/2

well that's one, what angle does that correspond to?
and what is the other intersection?

45 degrees

and 60 degrees

yes, 45 and 60, but you better use radians in calc|dw:1352754013710:dw|

|dw:1352754159926:dw|

so which fuunction is the outer radius and which one the inner radius for\[0\le\theta\le\frac\pi4\]?

r=2 is inner and r=2sqrt2 is outer...

yes, good :)
and for the interval\[\frac\pi4\le\theta\le\frac\pi3\]what is it?

r=4cos (theta) is outer and r=2sqrt2 is inner...

I think you have the inner wrong|dw:1352754952806:dw|

oh, inner is r=2

right
and what are your integrands?

I'm not too sure about that

you are integrating xy
sub in x=rcost, y=rsint

above t=theta in case you didn't guess
also you need to use dA for polar coordinate

so it'll be the double integral of r^2 cost sint...

that is the xy part, and what is dA in polar?

dr dt

sorry, r dr dt

yes, forget that extra r and you will be sorry :P
so can you write your integrals yet?

I dont think so

do you see that it will be two integrals?

im getting three rs and thethas

r=2 to 2√2 and t= 0 to pi/4
and
r=2√2 to 4cosθ and t= 0 to pi/3

Sorry about that. give me a sec

for the second integral: r=2 to 4cost and t= pi/4 to pi/3

good :)
so how 'bout setting up those integrals? (two double intergals)

that's the first one, yes

\[\int\limits_{\pi/4}^{\pi/3} \int\limits_{2}^{4\cos \Theta} r2\cosΘ\sinΘrdrdΘ\]

yes, good :)
then add them together to find the total voume
happy integrating

Oh, God this does not look like fun, but
Thanks for the help :)