## anonymous 3 years ago 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.

1. TuringTest

|dw:1352751755636:dw|

2. 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|

3. anonymous

1st octant...

4. anonymous

Make the equations equal to one another...

5. TuringTest

yes, and what do you get for theta?

6. anonymous

arccos 2^(1/2)/2

7. TuringTest

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

8. anonymous

45 degrees

9. anonymous

and 60 degrees

10. TuringTest

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

11. TuringTest

|dw:1352754159926:dw|

12. TuringTest

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

13. anonymous

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

14. TuringTest

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

15. anonymous

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

16. TuringTest

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

17. anonymous

oh, inner is r=2

18. TuringTest

right and what are your integrands?

19. anonymous

I'm not too sure about that

20. TuringTest

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

21. TuringTest

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

22. anonymous

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

23. TuringTest

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

24. anonymous

dr dt

25. anonymous

sorry, r dr dt

26. TuringTest

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

27. anonymous

I dont think so

28. TuringTest

do you see that it will be two integrals?

29. anonymous

im getting three rs and thethas

30. 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.

31. anonymous

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

32. 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

33. anonymous

Sorry about that. give me a sec

34. anonymous

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

35. TuringTest

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

36. anonymous

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

37. TuringTest

that's the first one, yes

38. anonymous

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

39. TuringTest

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

40. anonymous

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

41. 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!