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Using a polar double integral, find the volume in the first octant below the surface z=xy, inside cylinder r=4cos(theta), and between the cylinders x^2 + y^2=4 and x^2 + y^2=8.

Mathematics
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|dw:1352751755636:dw|
in polar the three graphs are\[r=2\]\[r=2\sqrt2\]\[r=4\cos\theta\]can you find the intersection points in terms of theta?|dw:1352752221119:dw|
1st octant...

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Other answers:

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
there are two regions to consider|dw:1352756391639:dw|what are the bounds for r and theta on the first integral? what are they on the second? those will be the bounds of your integral.
r=2 to 2√2 and t= 0 to pi/4 and r=2√2 to 4cosθ and t= 0 to pi/3
on the second region the lower bound on r is not 2√2 as we have already discussed, and the lower bound on theta is not 0, but the upper bound of the previous integral
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)
\[\int\limits_{0}^{\pi/4}\int\limits\limits_{2}^{2\sqrt{2}}r ^{2}\cos \Theta \sin \Theta r dr d \Theta \]
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 :)
It's really not that hard if you can just remember to integrate with respect to theta and r separately, but I'm still too lazy to do it. Welcome!

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