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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cross 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
 5 years ago
Best ResponseYou've already chosen the best response.0i got the equations but i cant figure out the positive negative problem with the square root

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0arman, you can solve this using cylindrical shells. \[\delta V = 2\pi (radius) (height)(elemental.thickness)\]\[\delta V = 2\pi x \left( \sqrt{r^2(x3/2)^2}\left( \sqrt{r^2(x3/2)^2} \right) \right) \delta x\]\[\delta V = 2\pi x \left( 2\sqrt{r^2(x3/2)^2}\right) \delta x\]where r = 3/4 (0.75)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your limits of integration would be x=3/4 to x=9/4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dang yo i just reposted this up top

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think im supposed to use washers though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, that can be done too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0at the end he has us compare our answer via washer to the answer via cylindes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok, just let me do it on paper.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm going to let a=3/2 and r=3/4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yesss, thank you so much

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\delta V = \pi \left[ \left( a+\sqrt{r^2y^2} \right)^2\left( a\sqrt{r^2y^2} \right)^2 \right]\delta y\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did you get something like that at first?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wow... i know my problem now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope nvrmnd i still dont.. yeah thats what i got

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, well all you have to do is expand that and integrate. \[\delta V=\pi \left( 4a \sqrt{r^2y^2} \right)\delta y\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i integrated then expanded...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[V=4a \pi \int\limits_{r}^{r}\sqrt{r^2y^2}dy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x and y are interchangeable (technically speaking) in this case?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, you're integrating over the interval on the yaxis.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, 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
 5 years ago
Best ResponseYou've already chosen the best response.0well...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 xaxis, 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
 5 years ago
Best ResponseYou've already chosen the best response.0I think I know where you were having problems too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you sub you get a negative numberunder the radical

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you integrate that expression for y (indefinite integral), you'll derive: \[\int\limits{}_{}\sqrt{r^2y^2}dy=\frac{1}{2}\left[ y \sqrt{r^2y^2}+r^2 \tan^{1}\left( \frac{y}{\sqrt{r^2y^2}} \right) \right]+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0plus constant (didn't show up).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0arctan is wayyy too complex for this O.o

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the directions request that it be rotated around the x axis >.<

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now...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
 5 years ago
Best ResponseYou've already chosen the best response.0You need to draw me a picture of how you're supposed to orient it. The volume you get will be the same.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can just switch axes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The volume I get is \[2\pi^2ar^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which is the volume of a torus.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this for class or fun?!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it can be both you know lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0what you got with the 2pi^2ar^2 is correct, hat much i know

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...that's no problem...just switch the x's for y's.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I rotated around the yaxis, but if you consider the geometry, by switching labels, 'suddenly' you're revolving the torus around the xaxis.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i understand that... i dont understand why my equations are wrong ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, your setup is right, since we both agreed, so it must be the algebra in between.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i should be using: v= int(pi(root(r2x2)+1.5)  int(pi(root(r2x2)+1.5))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's the same as what we did above. I used y. You can use x. Whereas I put my circle on the xaxis to rotate, you're putting it on the yaxis. 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
 5 years ago
Best ResponseYou've already chosen the best response.0You know, I will set the problem up with a circle on the yaxis. You can punch your teacher for me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol 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
 5 years ago
Best ResponseYou've already chosen the best response.0This might surprise you, but I got the same formula like I did for y, except now it has x's...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, you have your circle on the yaxis and it's centered at (0,a).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The circle's equation is \[x^2+(ya)^2=r^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0haha funny almost like i didnt know you can switch x and y . armaan not stupid ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solving the equation of the circle for those yvalues gives\[y_2=a+\sqrt{r^2x^2}\]\[y_1=a\sqrt{r^2x^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The area of the punctured disc is\[\pi(y_2^2y_1^2)\]\[=\pi [(a+\sqrt{r^2x^2})^2(a\sqrt{r^2x^2})^2]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The volume will just be the area multiplied by the 'width', delta x:\[\delta V = \pi[(a+\sqrt{r^2x^2})^2(a\sqrt{r^2x^2})^2] \delta x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From here on in it is algebra and integration. The problem's been set up.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The bit in the brackets:\[a^2+2\sqrt{r^2x^2}+r^2x^2(a^22\sqrt{r^2x^2}+r^2x^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry for causing the trouble on this :/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, after you punch your teacher, punch yourself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0brackets continued...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[4a \sqrt{r^2x^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When I expanded above, I forgot the 'a' next to the sqrt expressions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[a^2+2a \sqrt{r^2x^2}+r^2x^2(a^22a \sqrt{r^2x^2}+r^2x^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, so the volume is now,\[V=4 \pi a \int\limits_{r}^{r}\sqrt{r^2x^2}dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But you can recognize the integral as the area of a semicircle of radius r, which would be\[\pi \frac{r^2}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which means your volume is\[V=4\pi a \times \pi \frac{r^2}{2}=2 \pi^2 a r^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i see, but then why does it not work if you just subtract the original integrals without combining or simplifying

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You mean without doing the algebraic expansion way up there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But looking at it should make you go, "No way, I'm expanding" because you have radicals and squares.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0iight thanks, id say i feel retarded, but given what im trying to learn, i feel pretty normal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, you're doing fine. just need to calm down.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There aren't many people (proportionately speaking) who can understand this stuff, so you're doing exceptionally well.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0id say thats a bit of exaggeration haha i seriously want to sit in your shoes though, given what you know

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how much time did it take you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0depends...maths is so large, you're learning every day.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so thats what your research is for?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0= money = food + accommodation + clothes...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you have a math based job?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I'm teaching it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i see, you spend your time at the college either researching or lecturing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was doing some research into biophysics and something called Takahashi's problem. It's a mathematical physics problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you intend to solve or something of that sort the problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll be round for a while if you need anything else.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i need to go soon DX its 235 here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0iight i have no idea how you got 4a[etcetc]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can understand how, but how did you know it was 4a without actually expanding it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did expand it. The 4a comes from the fact you have \[2a \sqrt{r^2x^2}+2a \sqrt{r^2x^2}=4a \sqrt{r^2x^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok, i thought there was some hint or trick to it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No...you should probably go to bed.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0after i finish my doughnut .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0iight im done, so ima head to bed, thanks for the help, as always, gnight, ttyl

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and idk about the punching the teacher... hes pretty cool
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