## anonymous one year ago MEDAL!! Please help...! use triple integration to find the volume of the solid bounded by the cylinders x^2+y^2=9 and y^2+z^2=9

1. amistre64

well, what part are you having an issue with?

2. anonymous

the limit points im using if -3 to 3, -squareroot (9-x^2), squareroot (9-x^2), -squareroot (9-y^2), -squareroot (9-y^2)

3. anonymous

can you guide the first couple steps, im really confused please

4. amistre64

trying to see if i can see the boundary |dw:1436633020633:dw|

5. anonymous

is it the inner part of cylinder we are looking for?

6. amistre64

are your limits valid? other than that it really is just a mechanical process.

7. anonymous

there valid but im having trouble with the integration

8. amistre64

well lets see what we can work then dz from -sqrt(9-y^2) to sqrt(9-y^2) dy from -sqrt(9-x^2) to sqrt(9-x^2) dx from -3 to 3 you agree that this is how we want to move?

9. anonymous

yes

10. amistre64

then we simply work it from the inside out $\int \left(\int\left[\int dz\right]~dy\right)~dx$ $\int \left(\int z_b-z_a~dy\right)~dx$ to start with

11. amistre64

$\int \left(2\int \sqrt{9-y^2}~dy\right)~dx$ let y = 3sin(t) , dy = 3cos(t) dt $\int \left(2\int \sqrt{9-(3sin(t))^2}~3cos(t)~dt\right)~dx$ $\int \left(2\int \sqrt{9(1-sin^2(t))}~3cos(t)~dt\right)~dx$ $\int \left(2\int 3cos^2(t)*3cos(t)~dt\right)~dx$ $\int \left(18\int cos^3(t)~dt\right)~dx$ etc ..

12. anonymous

where did the sin^2t and cos come from

13. amistre64

i have a small error ... but its just using a trig substitution we let y = 3sin(t) and substitute all the y parts. sqrt(1-sin^2) = cos ... not cos^2 tho

14. anonymous

did you do squareroot (9-y^2) ==> squareroot (9-((9-x^2)^(1/2))?

15. anonymous

where did you plug in the limits for y?

16. amistre64

no sqrt(9 - y^2) dy if we let y = 3sin(t), the derivative allows us to replace dy by dt parts dy = 3cos(t) dt by replacing the appropriate parts ... sqrt(9 - (3sin(t))^2) * 3cos(t) dt

17. amistre64

my only concern for this is your limits ... im just not sure that the intersection of 2 cylindars is a sphere ... and that is what your limits are suggesting.

18. anonymous

right i see that...so my limits for dz are wrong?

19. amistre64

i believe your limits are bad ... im just not sure at the moment how we would correct them off hand. the draw tool is not that good at 3d imaging for me

20. anonymous

is there a formula... in mathetica it demonstrated 2 cylinders crossing eachother like a cross and what i was told to look for the volume of the non-intersecting upper part

21. amistre64

uh, you originally asked for the volume of the intersection ...

22. amistre64

|dw:1436634359956:dw| so you want the volume of something like this?

23. anonymous

24. anonymous

it looks like this

25. amistre64

right, and you want the volume of .. what now?

26. amistre64

if its just 2 cylndars intersecting, i thing the profile of the intersection is a "pringle" chip .. or a saddle back

27. amistre64

|dw:1436634627932:dw|

28. anonymous

yes ignore the red cylinder

29. anonymous

exactly like that

30. anonymous

from that part, how do you find the limits?

31. amistre64

now, you want to find the volume of the top part? or of the intersecting part? if its the top part, then we need to have a plane of something that cuts it off from going to infinity

32. anonymous

interesting part

33. amistre64

ok, interSECTing part :) yeah z limits are fine, -sqrt(9-y^2) to sqrt(9-y^2) x limits are fine, -3 to 3 its your y limits that need refined if anything

34. anonymous

good to hear its not completey wrong. y limits how to find them?

35. amistre64

still working on that ....

36. anonymous

its oki ! im still here

37. amistre64

https://www.youtube.com/watch?v=rASKzEbMXIQ this might be a bit more useful then me trying to remember things long dead past :) at least for a review

38. anonymous

thank you so much

39. anonymous

would you happen to remember greens thm?

40. amistre64

i never really covered greens thrm in my courses. so no.

41. anonymous

its oki..would you be staying online today? if its oki to ask you other questions

42. amistre64

im not going to be online much today .. got family stuff to do. but after the youtube video i do see where the issue was :) it was in the limits

43. anonymous

oh so the set up is the same?

44. amistre64

the setup is the same, but we need to better define the limits ... and using symmetry helps x^2+y^2=9 y^2+z^2=9 they both have a y^2 in common so lets define z and x in terms of y x = 0 to sqrt(9-y^2) z = 0 to sqrt(9-y^2) y = 0 to 3 and this defines 1/8 of our volume dz dx dy is a good order of integration

45. amistre64

$\int_{y=0}^{y=3}\int_{x=0}^{x=\sqrt{9-y^2}}\int_{z=0}^{z=\sqrt{9-y^2}}~dz~dx~dy$

46. anonymous

oki can you show me the frst integrration please

47. amistre64

well, using your fundamental thrm of calculus ... $\int_{0}^{b}dz=F(b)-F(0)$

48. amistre64

what is the integration of 1?

49. anonymous

x

50. amistre64

well, z in this case, since we are integrating with respect to z but yeah so z(b) - z(0) = z(b)

51. anonymous

so with z...then we plug in the limits of x?

52. amistre64

limits of z ...

53. amistre64

$\int_{0}^{\sqrt{9-y^2}}dz=z_b-z_a=\sqrt{9-y^2}-0$

54. amistre64

now, what is the middle integral? $\int_{0}^{\sqrt{9-y^2}}\sqrt{9-y^2}~dx=??$

55. anonymous

$\int\limits_{0}^{3} (1/2)(\sqrt{9-y^2}y+9\sin^{-1} (y/3)$

56. amistre64

not quite sqrt(9-y^2) , with respect to x is just some constant ... so, what is the integration of a constant again?

57. anonymous

x^2/2

58. amistre64

no, try again.

59. amistre64

what was the integration of 1?

60. anonymous

z

61. amistre64

the respective variable yes ... in this case we are doing dx, so its an x

62. amistre64

for some constant k $\int_{0}^{b}dx=b(k)-0(k)$

63. amistre64

in this case, k=sqrt(9-y^2), and b=sqrt(9-y^2) leaving us: $\int_{0}^{3}9-y^2~dy$

64. anonymous

oh ok so this is our middle integral

65. amistre64

yes, we did the first, then the middle, and we are now left with the end

66. anonymous

67. amistre64

dunno, i havent gotten to the end of it yet

68. amistre64

i dont think 19 is it

69. anonymous

18

70. amistre64

what is: $\int_{0}^{3}9-y^2~dy?$ work the steps for me

71. anonymous

$9y-y$

72. anonymous

9y-y^3/3

73. amistre64

9y - y^3/3 , given that y=3 yes 27 - 27/3 = 27(2/3) = 18 but recall that we only worked 1/8 of it, so we need to mulitply this by 8 to get the total volume.

74. anonymous

144...can you remind me again why we work with 1/8

75. anonymous

the limits for x and z why are they the same?

76. amistre64

we dont have to, but due to the symmetry of the intersection, it makes life a little simpler for some.

77. amistre64

those are better explained in the youtube video i posted.

78. anonymous

oki i understand...so much more clear. thank you

79. amistre64

good luck :)