joannaj93
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.
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TuringTest
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|dw:1352751755636:dw|
TuringTest
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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|
Algebraic!
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1st octant...
joannaj93
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Make the equations equal to one another...
TuringTest
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yes, and what do you get for theta?
joannaj93
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arccos 2^(1/2)/2
TuringTest
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well that's one, what angle does that correspond to?
and what is the other intersection?
joannaj93
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45 degrees
joannaj93
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and 60 degrees
TuringTest
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yes, 45 and 60, but you better use radians in calc|dw:1352754013710:dw|
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|dw:1352754159926:dw|
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so which fuunction is the outer radius and which one the inner radius for\[0\le\theta\le\frac\pi4\]?
joannaj93
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r=2 is inner and r=2sqrt2 is outer...
TuringTest
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yes, good :)
and for the interval\[\frac\pi4\le\theta\le\frac\pi3\]what is it?
joannaj93
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r=4cos (theta) is outer and r=2sqrt2 is inner...
TuringTest
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I think you have the inner wrong|dw:1352754952806:dw|
joannaj93
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oh, inner is r=2
TuringTest
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right
and what are your integrands?
joannaj93
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I'm not too sure about that
TuringTest
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you are integrating xy
sub in x=rcost, y=rsint
TuringTest
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above t=theta in case you didn't guess
also you need to use dA for polar coordinate
joannaj93
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so it'll be the double integral of r^2 cost sint...
TuringTest
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that is the xy part, and what is dA in polar?
joannaj93
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dr dt
joannaj93
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sorry, r dr dt
TuringTest
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yes, forget that extra r and you will be sorry :P
so can you write your integrals yet?
joannaj93
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I dont think so
TuringTest
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do you see that it will be two integrals?
joannaj93
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im getting three rs and thethas
TuringTest
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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.
joannaj93
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r=2 to 2√2 and t= 0 to pi/4
and
r=2√2 to 4cosθ and t= 0 to pi/3
TuringTest
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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
joannaj93
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Sorry about that. give me a sec
joannaj93
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for the second integral: r=2 to 4cost and t= pi/4 to pi/3
TuringTest
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good :)
so how 'bout setting up those integrals? (two double intergals)
joannaj93
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\[\int\limits_{0}^{\pi/4}\int\limits\limits_{2}^{2\sqrt{2}}r ^{2}\cos \Theta \sin \Theta r dr d \Theta \]
TuringTest
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that's the first one, yes
joannaj93
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\[\int\limits_{\pi/4}^{\pi/3} \int\limits_{2}^{4\cos \Theta} r2\cosΘ\sinΘrdrdΘ\]
TuringTest
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yes, good :)
then add them together to find the total voume
happy integrating
joannaj93
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Oh, God this does not look like fun, but
Thanks for the help :)
TuringTest
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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!