anonymous
  • anonymous
volume of a torus...?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • anonymous
i got the equations but i cant figure out the positive negative problem with the square root
anonymous
  • 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)

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anonymous
  • anonymous
Your limits of integration would be x=3/4 to x=9/4.
anonymous
  • anonymous
dang yo i just reposted this up top
anonymous
  • anonymous
i think im supposed to use washers though
anonymous
  • anonymous
oh
anonymous
  • anonymous
Well, that can be done too.
anonymous
  • anonymous
at the end he has us compare our answer via washer to the answer via cylindes
anonymous
  • 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
  • anonymous
ok, just let me do it on paper.
anonymous
  • anonymous
Okay, ready?
anonymous
  • anonymous
I'm going to let a=3/2 and r=3/4.
anonymous
  • anonymous
yesss, thank you so much
anonymous
  • 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
  • anonymous
Did you get something like that at first?
anonymous
  • anonymous
oh wow... i know my problem now
anonymous
  • anonymous
nope nvrmnd i still dont.. yeah thats what i got
anonymous
  • 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
  • anonymous
i integrated then expanded...
anonymous
  • anonymous
\[V=4a \pi \int\limits_{-r}^{r}\sqrt{r^2-y^2}dy\]
anonymous
  • anonymous
x and y are interchangeable (technically speaking) in this case?
anonymous
  • anonymous
No, you're integrating over the interval on the y-axis.
anonymous
  • 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
  • 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
  • anonymous
I think I know where you were having problems too.
anonymous
  • anonymous
when you sub you get a negative numberunder the radical
anonymous
  • anonymous
or 0 in this case
anonymous
  • 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
  • anonymous
plus constant (didn't show up).
anonymous
  • anonymous
arctan is wayyy too complex for this O.o
anonymous
  • anonymous
and the directions request that it be rotated around the x axis >.<
anonymous
  • 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
  • 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
  • anonymous
You can just switch axes.
anonymous
  • anonymous
The volume I get is \[2\pi^2ar^2\]
anonymous
  • anonymous
which is the volume of a torus.
anonymous
  • anonymous
Is this for class or fun?!
anonymous
  • anonymous
it can be both you know lol
anonymous
  • 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
  • anonymous
what you got with the 2pi^2ar^2 is correct, hat much i know
anonymous
  • anonymous
Okay...that's no problem...just switch the x's for y's.
anonymous
  • 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
  • anonymous
i understand that... i dont understand why my equations are wrong ><
anonymous
  • anonymous
Well, your setup is right, since we both agreed, so it must be the algebra in between.
anonymous
  • anonymous
i should be using: v= int(pi(root(r2-x2)+1.5) - int(pi(-root(r2-x2)+1.5))
anonymous
  • 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
  • anonymous
You know, I will set the problem up with a circle on the y-axis. You can punch your teacher for me.
anonymous
  • 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
  • anonymous
This might surprise you, but I got the same formula like I did for y, except now it has x's...
anonymous
  • anonymous
Okay, you have your circle on the y-axis and it's centered at (0,a).
anonymous
  • anonymous
The circle's equation is \[x^2+(y-a)^2=r^2\]
anonymous
  • 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
  • anonymous
haha funny almost like i didnt know you can switch x and y -.- armaan not stupid ><
anonymous
  • 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
  • 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
  • anonymous
right
anonymous
  • anonymous
happy?
anonymous
  • anonymous
this makes sense
anonymous
  • 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
  • anonymous
From here on in it is algebra and integration. The problem's been set up.
anonymous
  • 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
  • anonymous
sorry for causing the trouble on this :/
anonymous
  • anonymous
yeah, after you punch your teacher, punch yourself
anonymous
  • anonymous
:p
anonymous
  • anonymous
brackets continued...
anonymous
  • anonymous
haha well said
anonymous
  • anonymous
\[4a \sqrt{r^2-x^2}\]
anonymous
  • anonymous
When I expanded above, I forgot the 'a' next to the sqrt expressions.
anonymous
  • 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
  • anonymous
should be this ^
anonymous
  • anonymous
Okay, so the volume is now,\[V=4 \pi a \int\limits_{-r}^{r}\sqrt{r^2-x^2}dx\]
anonymous
  • 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
  • anonymous
which means your volume is\[V=4\pi a \times \pi \frac{r^2}{2}=2 \pi^2 a r^2\]
anonymous
  • anonymous
Okay?
anonymous
  • anonymous
ok i see, but then why does it not work if you just subtract the original integrals without combining or simplifying
anonymous
  • anonymous
You mean without doing the algebraic expansion way up there?
anonymous
  • anonymous
It should.
anonymous
  • anonymous
But looking at it should make you go, "No way, I'm expanding" because you have radicals and squares.
anonymous
  • anonymous
iight thanks, id say i feel retarded, but given what im trying to learn, i feel pretty normal
anonymous
  • anonymous
yeah, you're doing fine. just need to calm down.
anonymous
  • anonymous
There aren't many people (proportionately speaking) who can understand this stuff, so you're doing exceptionally well.
anonymous
  • anonymous
id say thats a bit of exaggeration haha i seriously want to sit in your shoes though, given what you know
anonymous
  • anonymous
in time...
anonymous
  • anonymous
how much time did it take you?
anonymous
  • anonymous
depends...maths is so large, you're learning every day.
anonymous
  • anonymous
do you have a phD?
anonymous
  • anonymous
Not yet.
anonymous
  • anonymous
so thats what your research is for?
anonymous
  • anonymous
pretty much + work!
anonymous
  • anonymous
= money = food + accommodation + clothes...
anonymous
  • anonymous
do you have a math based job?
anonymous
  • anonymous
Yes, I'm teaching it.
anonymous
  • anonymous
i see, you spend your time at the college either researching or lecturing
anonymous
  • anonymous
what is your phD on?
anonymous
  • anonymous
I was doing some research into biophysics and something called Takahashi's problem. It's a mathematical physics problem.
anonymous
  • anonymous
do you intend to solve or something of that sort the problem?
anonymous
  • anonymous
sorry?
anonymous
  • anonymous
yes
anonymous
  • anonymous
but we'll see.
anonymous
  • anonymous
I'll be round for a while if you need anything else.
anonymous
  • anonymous
i need to go soon DX its 235 here
anonymous
  • anonymous
iight i have no idea how you got 4a[etcetc]
anonymous
  • anonymous
i can understand how, but how did you know it was 4a without actually expanding it?
anonymous
  • 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
  • anonymous
oh ok, i thought there was some hint or trick to it
anonymous
  • anonymous
No...you should probably go to bed.
anonymous
  • anonymous
after i finish my doughnut -.-
anonymous
  • anonymous
iight im done, so ima head to bed, thanks for the help, as always, gnight, ttyl
anonymous
  • anonymous
and idk about the punching the teacher... hes pretty cool

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