Volume of toriodal region (engineering related)

[blackbelt5400]

Assume that 0<r<R.

Consider the circle $$x^2+(y+R)^2=r^2$$. Obviously, if we revolve this circle about the x-axis, we get a torus whose volume is $$(\pi r^2)(2\pi R)$$.

Now consider a point $$y_1$$ such that R-r<$$y_1$$<R. and let $$(x_1,y_1)$$ be a point on the circle. The volume of the solid formed by revolving the region inside of the circle and below $$y_1$$ about the x-axis is

$$2\pi [(R^2+r^2-y_1^2)x_1+R(x_1\sqrt{r^2-x_1^2}+r^2\sin^{-1}{(\frac{x_1}{r})})-\frac{x_1^3}{3}]$$.

Assume that a storage tank determined by this solid is filled to a height "h". What is V(h)?

I have attempted this in rectangular coordinates, but I don't believe that it's possible to set up an integral with these weird bounds. Changing to shperical or cylindrical coordinates will not work because this would introduce an error when integrating the distance from the origin.

I was told to try toroidal coordinates $$(\sigma, \tau, \phi)$$, and I have solved for the foci and the constant $$\tau$$ for this problem, but I don't know how to set up the integral.

 

[blackbelt5400]

This problem seems to be ridiculously hard in this form, so I've considered that it might be useful to try to find the volume of an entire torus as a function of height. If we want h>0, then the torus must be shifted up so that it is tangential to the xy-plane.

Manipulating the equation of a torus from Wolfram (http://mathworld.wolfram.com/Torus.html), the equation of a torus oriented in the xz direction and translated up a distance of c+a would be

$$(c-\sqrt{x^2+(z-c-a)^2})^2+y^2=a^2$$,

where "a" is the radius of the tube and "c" in the distance from the center of the torus to the center of the tube.

Finding the volume as a function of height will require a triple integral, and establishing the bounds correctly on the coordintes is essential to approaching this. Since we want to integrate as a function of height, we'll integrate z last from 0 to h. This means solving the equation for x and y first. It seems most logical to start with y. We get:

$$y= \pm \sqrt{a^2-(c-\sqrt{x^2+(z-c-a)^2})^2}$$

By symmetry, we can integrate y from 0 to the positive form of this expression and double the result.

We now look to the projection of the torus on the xz-plane in order to find the bounds on x. We have the equation:

$$(c-\sqrt{x^2+(z-c-a)^2})^2=a^2$$

Solving for x yields:

$$x= \pm \sqrt{(c\mp a)^2-(z-c-a)^2}$$

Again, by symmetry, I believe that we can integrate x from 0 to the strictly positive values of this expression and multiply the result by 4. Combining these yields:

$$V(h) = 8 \int_0^h\int_0^{\sqrt{(c+a)^2-(z-c-a)^2}}\int_0^{\sqrt{a^2-(c-\sqrt{x^2+(z-c-a)^2})^2}} dy dx dz$$

Integrating y is as far as I can get (which is trivial), and Maple 11 isn't any help. I'm stuck at

$$V(h) = 8 \int_0^h\int_0^{\sqrt{(c+a)^2-(z-c-a)^2}}{\sqrt{a^2-(c-\sqrt{x^2+(z-c-a)^2})^2}} dx dz$$

Can anyone help?

 

[danong]

Hi there,
I'm a little bit unclear with what you mean by
" The volume of the solid formed by revolving the region inside of the circle and below about the x-axis ",

I'm working hard on this question,
however,
my solution form seems to be different from your,
so i was wondering if i had misunderstood to the revolving issue.

so, could you somehow sketch a graph about it? a simple one will do,
thanks.

Have a nice day,

Regards,
Daniel.

 

[chiro]

This problem seems to be ridiculously hard in this form, so I've considered that it might be useful to try to find the volume of an entire torus as a function of height. If we want h>0, then the torus must be shifted up so that it is tangential to the xy-plane.

Manipulating the equation of a torus from Wolfram (http://mathworld.wolfram.com/Torus.html), the equation of a torus oriented in the xz direction and translated up a distance of c+a would be

$$(c-\sqrt{x^2+(z-c-a)^2})^2+y^2=a^2$$,

where "a" is the radius of the tube and "c" in the distance from the center of the torus to the center of the tube.

Finding the volume as a function of height will require a triple integral, and establishing the bounds correctly on the coordintes is essential to approaching this. Since we want to integrate as a function of height, we'll integrate z last from 0 to h. This means solving the equation for x and y first. It seems most logical to start with y. We get:

$$y= \pm \sqrt{a^2-(c-\sqrt{x^2+(z-c-a)^2})^2}$$

By symmetry, we can integrate y from 0 to the positive form of this expression and double the result.

We now look to the projection of the torus on the xz-plane in order to find the bounds on x. We have the equation:

$$(c-\sqrt{x^2+(z-c-a)^2})^2=a^2$$

Solving for x yields:

$$x= \pm \sqrt{(c\mp a)^2-(z-c-a)^2}$$

Again, by symmetry, I believe that we can integrate x from 0 to the strictly positive values of this expression and multiply the result by 4. Combining these yields:

$$V(h) = 8 \int_0^h\int_0^{\sqrt{(c+a)^2-(z-c-a)^2}}\int_0^{\sqrt{a^2-(c-\sqrt{x^2+(z-c-a)^2})^2}} dy dx dz$$

Integrating y is as far as I can get (which is trivial), and Maple 11 isn't any help. I'm stuck at

$$V(h) = 8 \int_0^h\int_0^{\sqrt{(c+a)^2-(z-c-a)^2}}{\sqrt{a^2-(c-\sqrt{x^2+(z-c-a)^2})^2}} dx dz$$

Can anyone help?

Have you tried to setup a numerical integration? I'm not sure what transforms you can use analytically but surely a numerical approximation would exist to that function?

You mentioned using maple. I'm a little familiar with it but have you declared the function in maple and then done through numeric analysis found a solution?

Theres also an expansion for the square root of a function as a taylor series expansion. Its possible you could use that and then get the answer to within a given accuracy in so many digits. Check something like mathworld or wiki or some textbook or math repository for the mclaurin series expansion. At least with the series expansion you might get something similar though just looking at I would try something numeric to start with and see what you get.

 

[blackbelt5400]

Hi there,
I'm a little bit unclear with what you mean by
" The volume of the solid formed by revolving the region inside of the circle and below about the x-axis ",

Actually, what I said was "revolving the region inside of the circle and below $$y_1$$ about the x-axis "

I've attached a picture showing the shaded region that is revolved about the x-axis. Just use the washer method, using the outer radius R(x) as the circle and the inner radius r(x) as the line y=$$y_1$$. Then

$$V=2\pi \int_0^{x_1}[R(x)]^2-[r(x)]^2 dx$$

 

[blackbelt5400]

Have you tried to setup a numerical integration? I'm not sure what transforms you can use analytically but surely a numerical approximation would exist to that function?

You mentioned using maple. I'm a little familiar with it but have you declared the function in maple and then done through numeric analysis found a solution?

Theres also an expansion for the square root of a function as a taylor series expansion. Its possible you could use that and then get the answer to within a given accuracy in so many digits. Check something like mathworld or wiki or some textbook or math repository for the mclaurin series expansion. At least with the series expansion you might get something similar though just looking at I would try something numeric to start with and see what you get.

The purpose of the paper I'm writing is to establish exact formulas for the volume in terms of height. The current approximations used in these cases are somewhat inadequate and I'm trying to abolish them completely.

 

[danong]

Actually, what I said was "revolving the region inside of the circle and below $$y_1$$ about the x-axis "

I've attached a picture showing the shaded region that is revolved about the x-axis. Just use the washer method, using the outer radius R(x) as the circle and the inner radius r(x) as the line y=$$y_1$$. Then

$$V=2\pi \int_0^{x_1}[R(x)]^2-[r(x)]^2 dx$$

Oh thanks for your graph it's clear and pretty.
Anyway,
i guess it is the one i had in my paper as well,
i had solved it yesterday when i last replied,
however my solution is a different approach,
so i wonder if you would like to have a look.

Can i work my draft paper into a computer paper tommorow?
as i am having my another paper work right now,
i will let you know when i am published together with my working.

Thanks =)

 

[danong]

sorry for my late reply,
i have been delayed due to some work submission,
btw i had finally make it,
any correction or mistake please inform me,
thanks & have a nice day.



Regards,
Daniel.

 

[blackbelt5400]

sorry for my late reply,
i have been delayed due to some work submission,
btw i had finally make it,
any correction or mistake please inform me,
thanks & have a nice day.



Regards,
Daniel.

I like your approach. I haven't looked it over yet, but it appears quite promising. However, I would like to comment that your picture 4.0a shows you calculating the "inside" toroidal region. The problem my paper considers is the "outside". However, I think only a slight change to you method is all that is required. I'll look over it this week.

Thanks,
Kenny