## anonymous 5 years ago volume of a torus...?

1. anonymous

cross section of the circle has a radius of .75 and radius from center of torus to center of cross section circle is 1.5

2. anonymous

i got the equations but i cant figure out the positive negative problem with the square root

3. anonymous

arman, you can solve this using cylindrical shells. $\delta V = 2\pi (radius) (height)(elemental.thickness)$$\delta V = 2\pi x \left( \sqrt{r^2-(x-3/2)^2}-\left( -\sqrt{r^2-(x-3/2)^2} \right) \right) \delta x$$\delta V = 2\pi x \left( 2\sqrt{r^2-(x-3/2)^2}\right) \delta x$where r = 3/4 (0.75)

4. anonymous

Your limits of integration would be x=3/4 to x=9/4.

5. anonymous

dang yo i just reposted this up top

6. anonymous

i think im supposed to use washers though

7. anonymous

oh

8. anonymous

Well, that can be done too.

9. anonymous

at the end he has us compare our answer via washer to the answer via cylindes

10. anonymous

but how?? if you can just tell me what integrals to use and why i can figure it out but i keep having my areas cancell out of i have a root of a negative

11. anonymous

ok, just let me do it on paper.

12. anonymous

13. anonymous

I'm going to let a=3/2 and r=3/4.

14. anonymous

yesss, thank you so much

15. anonymous

$\delta V = \pi \left[ \left( a+\sqrt{r^2-y^2} \right)^2-\left( a-\sqrt{r^2-y^2} \right)^2 \right]\delta y$

16. anonymous

Did you get something like that at first?

17. anonymous

oh wow... i know my problem now

18. anonymous

nope nvrmnd i still dont.. yeah thats what i got

19. anonymous

Okay, well all you have to do is expand that and integrate. $\delta V=\pi \left( 4a \sqrt{r^2-y^2} \right)\delta y$

20. anonymous

i integrated then expanded...

21. anonymous

$V=4a \pi \int\limits_{-r}^{r}\sqrt{r^2-y^2}dy$

22. anonymous

x and y are interchangeable (technically speaking) in this case?

23. anonymous

No, you're integrating over the interval on the y-axis.

24. anonymous

right, but like as long as i sub for that variable, it doesn't matter if you write x or y physical interpretation will rotate obviously)?

25. anonymous

well...if you keep to the understanding of what x is in this situation and you stick to the same limits of integration. Otherwise, if you want to integrate over the x-axis, you're going to have to make a substitution for y in terms of x, and then find the differential of y with respect to x. It's more work.

26. anonymous

I think I know where you were having problems too.

27. anonymous

when you sub you get a negative numberunder the radical

28. anonymous

or 0 in this case

29. anonymous

If you integrate that expression for y (indefinite integral), you'll derive: $\int\limits{}_{}\sqrt{r^2-y^2}dy=\frac{1}{2}\left[ y \sqrt{r^2-y^2}+r^2 \tan^{-1}\left( \frac{y}{\sqrt{r^2-y^2}} \right) \right]+c$

30. anonymous

plus constant (didn't show up).

31. anonymous

arctan is wayyy too complex for this O.o

32. anonymous

and the directions request that it be rotated around the x axis >.<

33. anonymous

Now...when you approach the limit r from within your interval, you are going to approach zero in the radical from the positive side...yes, the denominator will approach zero, by arctan approach positive pi/2. When y approaches -r, again, the denom. will approach zero, but arctan will approach NEGATIVE pi/2. When you subtract (i.e. limits), you'll be adding 2 lots of pi/2...there'll be no cancelling.

34. anonymous

You need to draw me a picture of how you're supposed to orient it. The volume you get will be the same.

35. anonymous

You can just switch axes.

36. anonymous

The volume I get is $2\pi^2ar^2$

37. anonymous

which is the volume of a torus.

38. anonymous

Is this for class or fun?!

39. anonymous

it can be both you know lol

40. anonymous

i dont know anymore than the following: measure your torus: radius of circular cross section of the "tube" = a = _____ (i put .75in) radius rom center of torus to center of tube= b =______ (i put 1.5in) write the integral that calculates the volume of the torus by revolving the circle around the x/axis. show steps. show what A(x) is.

41. anonymous

what you got with the 2pi^2ar^2 is correct, hat much i know

42. anonymous

Okay...that's no problem...just switch the x's for y's.

43. anonymous

I rotated around the y-axis, but if you consider the geometry, by switching labels, 'suddenly' you're revolving the torus around the x-axis.

44. anonymous

i understand that... i dont understand why my equations are wrong ><

45. anonymous

Well, your setup is right, since we both agreed, so it must be the algebra in between.

46. anonymous

i should be using: v= int(pi(root(r2-x2)+1.5) - int(pi(-root(r2-x2)+1.5))

47. anonymous

That's the same as what we did above. I used y. You can use x. Whereas I put my circle on the x-axis to rotate, you're putting it on the y-axis. All you need to do is switch y for x in the formulas above. You're not changing the mathematics because, in this context, x and y and dummy variables.

48. anonymous

You know, I will set the problem up with a circle on the y-axis. You can punch your teacher for me.

49. anonymous

lol i completely agree with everything youve said i just dont know why you got a different answer, ill try it another dozen times and repost if i still fail miserably

50. anonymous

This might surprise you, but I got the same formula like I did for y, except now it has x's...

51. anonymous

Okay, you have your circle on the y-axis and it's centered at (0,a).

52. anonymous

The circle's equation is $x^2+(y-a)^2=r^2$

53. anonymous

The washer method says that the area of the washer is that of a punctured disc; that is, $\pi y_2^2- \pi y_1^2$

54. anonymous

haha funny almost like i didnt know you can switch x and y -.- armaan not stupid ><

55. anonymous

Solving the equation of the circle for those y-values gives$y_2=a+\sqrt{r^2-x^2}$$y_1=a-\sqrt{r^2-x^2}$

56. anonymous

The area of the punctured disc is$\pi(y_2^2-y_1^2)$$=\pi [(a+\sqrt{r^2-x^2})^2-(a-\sqrt{r^2-x^2})^2]$

57. anonymous

right

58. anonymous

happy?

59. anonymous

this makes sense

60. anonymous

The volume will just be the area multiplied by the 'width', delta x:$\delta V = \pi[(a+\sqrt{r^2-x^2})^2-(a-\sqrt{r^2-x^2})^2] \delta x$

61. anonymous

From here on in it is algebra and integration. The problem's been set up.

62. anonymous

The bit in the brackets:$a^2+2\sqrt{r^2-x^2}+r^2-x^2-(a^2-2\sqrt{r^2-x^2}+r^2-x^2)$

63. anonymous

sorry for causing the trouble on this :/

64. anonymous

yeah, after you punch your teacher, punch yourself

65. anonymous

:p

66. anonymous

brackets continued...

67. anonymous

haha well said

68. anonymous

$4a \sqrt{r^2-x^2}$

69. anonymous

When I expanded above, I forgot the 'a' next to the sqrt expressions.

70. anonymous

$a^2+2a \sqrt{r^2-x^2}+r^2-x^2-(a^2-2a \sqrt{r^2-x^2}+r^2-x^2)$

71. anonymous

should be this ^

72. anonymous

Okay, so the volume is now,$V=4 \pi a \int\limits_{-r}^{r}\sqrt{r^2-x^2}dx$

73. anonymous

But you can recognize the integral as the area of a semi-circle of radius r, which would be$\pi \frac{r^2}{2}$

74. anonymous

which means your volume is$V=4\pi a \times \pi \frac{r^2}{2}=2 \pi^2 a r^2$

75. anonymous

Okay?

76. anonymous

ok i see, but then why does it not work if you just subtract the original integrals without combining or simplifying

77. anonymous

You mean without doing the algebraic expansion way up there?

78. anonymous

It should.

79. anonymous

But looking at it should make you go, "No way, I'm expanding" because you have radicals and squares.

80. anonymous

iight thanks, id say i feel retarded, but given what im trying to learn, i feel pretty normal

81. anonymous

yeah, you're doing fine. just need to calm down.

82. anonymous

There aren't many people (proportionately speaking) who can understand this stuff, so you're doing exceptionally well.

83. anonymous

id say thats a bit of exaggeration haha i seriously want to sit in your shoes though, given what you know

84. anonymous

in time...

85. anonymous

how much time did it take you?

86. anonymous

depends...maths is so large, you're learning every day.

87. anonymous

do you have a phD?

88. anonymous

Not yet.

89. anonymous

so thats what your research is for?

90. anonymous

pretty much + work!

91. anonymous

= money = food + accommodation + clothes...

92. anonymous

do you have a math based job?

93. anonymous

Yes, I'm teaching it.

94. anonymous

i see, you spend your time at the college either researching or lecturing

95. anonymous

96. anonymous

I was doing some research into biophysics and something called Takahashi's problem. It's a mathematical physics problem.

97. anonymous

do you intend to solve or something of that sort the problem?

98. anonymous

sorry?

99. anonymous

yes

100. anonymous

but we'll see.

101. anonymous

I'll be round for a while if you need anything else.

102. anonymous

i need to go soon DX its 235 here

103. anonymous

iight i have no idea how you got 4a[etcetc]

104. anonymous

i can understand how, but how did you know it was 4a without actually expanding it?

105. anonymous

I did expand it. The 4a comes from the fact you have $2a \sqrt{r^2-x^2}+2a \sqrt{r^2-x^2}=4a \sqrt{r^2-x^2}$

106. anonymous

oh ok, i thought there was some hint or trick to it

107. anonymous

No...you should probably go to bed.

108. anonymous

after i finish my doughnut -.-

109. anonymous

iight im done, so ima head to bed, thanks for the help, as always, gnight, ttyl

110. anonymous

and idk about the punching the teacher... hes pretty cool