anonymous
  • anonymous
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
jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
|dw:1352751755636:dw|
TuringTest
  • 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
  • anonymous
1st octant...

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