Question

volume of a torus...?

Answer 1

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

Answer 2

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

Answer 3

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)

Answer 4

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

Answer 5

dang yo i just reposted this up top

Answer 6

i think im supposed to use washers though

Answer 7

oh

Answer 8

Well, that can be done too.

Answer 9

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

Answer 10

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

Answer 11

ok, just let me do it on paper.

Answer 12

Okay, ready?

Answer 13

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

Answer 14

yesss, thank you so much

Answer 15

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

Answer 16

Did you get something like that at first?

Answer 17

oh wow... i know my problem now

Answer 18

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

Answer 19

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

Answer 20

i integrated then expanded...

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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)?

Answer 25

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.

Answer 26

I think I know where you were having problems too.

Answer 27

when you sub you get a negative numberunder the radical

Answer 28

or 0 in this case

Answer 29

If you integrate that expression for y (indefinite integral), you'll derive:

\[\int\limits{}_{}\sqrt{r^2-y^2}dy=\frac{1}{2}\left[ y \sqrt{r^2-y^2}+r^2 \tan^{-1}\left( \frac{y}{\sqrt{r^2-y^2}} \right) \right]+c\]

Answer 30

plus constant (didn't show up).

Answer 31

arctan is wayyy too complex for this O.o

Answer 32

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

Answer 33

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.

Answer 34

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

Answer 35

You can just switch axes.

Answer 36

The volume I get is \[2\pi^2ar^2\]

Answer 37

which is the volume of a torus.

Answer 38

Is this for class or fun?!

Answer 39

it can be both you know lol

Answer 40

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.

Answer 41

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

Answer 42

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

Answer 43

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.

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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.

Answer 48

You know, I will set the problem up with a circle on the y-axis.

You can punch your teacher for me.

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

right

Answer 58

happy?

Answer 59

this makes sense

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

sorry for causing the trouble on this :/

Answer 64

yeah, after you punch your teacher, punch yourself

Answer 65

:p

Answer 66

brackets continued...

Answer 67

haha well said

Answer 68

\[4a \sqrt{r^2-x^2}\]

Answer 69

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

Answer 70

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

Answer 71

should be this ^

Answer 72

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

Answer 73

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

Answer 74

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

Answer 75

Okay?

Answer 76

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

Answer 77

You mean without doing the algebraic expansion way up there?

Answer 78

It should.

Answer 79

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

Answer 80

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

Answer 81

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

Answer 82

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

Answer 83

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

Answer 84

in time...

Answer 85

how much time did it take you?

Answer 86

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

Answer 87

do you have a phD?

Answer 88

Not yet.

Answer 89

so thats what your research is for?

Answer 90

pretty much + work!

Answer 91

= money = food + accommodation + clothes...

Answer 92

do you have a math based job?

Answer 93

Yes, I'm teaching it.

Answer 94

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

Answer 95

what is your phD on?

Answer 96

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

Answer 97

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

Answer 98

sorry?