volume of a torus...?

- anonymous

volume of a torus...?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

- anonymous

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

- 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)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

dang yo i just reposted this up top

- anonymous

i think im supposed to use washers though

- anonymous

oh

- anonymous

Well, that can be done too.

- anonymous

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

- 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

- anonymous

ok, just let me do it on paper.

- anonymous

Okay, ready?

- anonymous

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

- anonymous

yesss, thank you so much

- anonymous

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

- anonymous

Did you get something like that at first?

- anonymous

oh wow... i know my problem now

- anonymous

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

- anonymous

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

- anonymous

i integrated then expanded...

- anonymous

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

- anonymous

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

- anonymous

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

- 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)?

- 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.

- anonymous

I think I know where you were having problems too.

- anonymous

when you sub you get a negative numberunder the radical

- anonymous

or 0 in this case

- anonymous

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

- anonymous

plus constant (didn't show up).

- anonymous

arctan is wayyy too complex for this O.o

- anonymous

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

- 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.

- anonymous

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

- anonymous

You can just switch axes.

- anonymous

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

- anonymous

which is the volume of a torus.

- anonymous

Is this for class or fun?!

- anonymous

it can be both you know lol

- 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.

- anonymous

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

- anonymous

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

- 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.

- anonymous

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

- anonymous

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

- anonymous

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

- 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.

- anonymous

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

- 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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

right

- anonymous

happy?

- anonymous

this makes sense

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

sorry for causing the trouble on this :/

- anonymous

yeah, after you punch your teacher, punch yourself

- anonymous

:p

- anonymous

brackets continued...

- anonymous

haha well said

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

should be this ^

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Okay?

- anonymous

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

- anonymous

You mean without doing the algebraic expansion way up there?

- anonymous

It should.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

in time...

- anonymous

how much time did it take you?

- anonymous

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

- anonymous

do you have a phD?

- anonymous

Not yet.

- anonymous

so thats what your research is for?

- anonymous

pretty much + work!

- anonymous

= money = food + accommodation + clothes...

- anonymous

do you have a math based job?

- anonymous

Yes, I'm teaching it.

- anonymous

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

- anonymous

what is your phD on?

- anonymous

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

- anonymous

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

- anonymous

sorry?

- anonymous

yes

- anonymous

but we'll see.

- anonymous

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

- anonymous

i need to go soon DX its 235 here

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

No...you should probably go to bed.

- anonymous

after i finish my doughnut -.-

- anonymous

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.