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.

- anonymous

- jamiebookeater

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- TuringTest

|dw:1352751755636:dw|

- TuringTest

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|

- anonymous

1st octant...

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## More answers

- anonymous

Make the equations equal to one another...

- TuringTest

yes, and what do you get for theta?

- anonymous

arccos 2^(1/2)/2

- TuringTest

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

- anonymous

45 degrees

- anonymous

and 60 degrees

- TuringTest

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

- TuringTest

|dw:1352754159926:dw|

- TuringTest

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

- anonymous

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

- TuringTest

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

- anonymous

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

- TuringTest

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

- anonymous

oh, inner is r=2

- TuringTest

right
and what are your integrands?

- anonymous

I'm not too sure about that

- TuringTest

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

- TuringTest

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

- anonymous

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

- TuringTest

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

- anonymous

dr dt

- anonymous

sorry, r dr dt

- TuringTest

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

- anonymous

I dont think so

- TuringTest

do you see that it will be two integrals?

- anonymous

im getting three rs and thethas

- TuringTest

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.

- anonymous

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

- TuringTest

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

- anonymous

Sorry about that. give me a sec

- anonymous

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

- TuringTest

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

- anonymous

\[\int\limits_{0}^{\pi/4}\int\limits\limits_{2}^{2\sqrt{2}}r ^{2}\cos \Theta \sin \Theta r dr d \Theta \]

- TuringTest

that's the first one, yes

- anonymous

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

- TuringTest

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

- anonymous

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

- TuringTest

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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