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Solids of rotation(volume of a torus)

by nlsherrill
Tags: rotationvolume, solids, torus
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nlsherrill
#1
Jul13-10, 07:33 PM
P: 322
The problem reads(from Stewart Calculus Concepts and Contexts 4th edition, Ch.6 section 2 pg. 447 #45

a)Set up an integral for the volume of a solid torus(the donut-shaped solid shown in the figure) with radii r and R
b)By interpreting the integral as an area, find the volume of the torus




2. Relevant equations



3. The attempt at a solution

[tex]\int\\pi(1-(R+r))^2-\pi(1-(R-r)^2))dx[/tex]

according to the back of the book this isn't correct. I basically treated this like a washer and did the area of the outer-inner functions and integrated. Heres a picture from the book that may help..

[/b]
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Mark44
#2
Jul13-10, 08:36 PM
Mentor
P: 21,283
Quote Quote by nlsherrill View Post
The problem reads(from Stewart Calculus Concepts and Contexts 4th edition, Ch.6 section 2 pg. 447 #45

a)Set up an integral for the volume of a solid torus(the donut-shaped solid shown in the figure) with radii r and R
b)By interpreting the integral as an area, find the volume of the torus



2. Relevant equations



3. The attempt at a solution

[tex]\int\\pi(1-(R+r))^2-\pi(1-(R-r)^2))[/tex]

according to the back of the book this isn't correct. I basically treated this like a washer and did the area of the outer-inner functions and integrated. Heres a picture from the book that may help..
Your integral doesn't have dx or dy, or any indication of what the variable of integration is. Also, why does 1 appear in your integrand?

One approach is to use cylindrical shells of thickness [itex]\Delta x[/itex]. You will need to find the equation of the circle. Its center is at (R, 0) and its radius is r. Find an expression for the incremental area [itex]\Delta A[/itex]. I'm sure your text has an explanation of the shell method and several examples.
Gregg
#3
Jul13-10, 08:48 PM
P: 463
Can't you just treat this as a bent cylinder of length [itex] 2 \pi R [/itex] and base [itex]\pi r^2 [/itex]?

Mark44
#4
Jul13-10, 09:19 PM
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P: 21,283
Solids of rotation(volume of a torus)

I don't think so, plus that doesn't get you the answer in the back of the book.

Edit: Actually, it does work.
Gregg
#5
Jul13-10, 09:27 PM
P: 463
Isn't the volume of a torus [itex] 2\pi^2r^2 R [/itex]?
nlsherrill
#6
Jul13-10, 09:29 PM
P: 322
Quote Quote by Mark44 View Post
Your integral doesn't have dx or dy, or any indication of what the variable of integration is. Also, why does 1 appear in your integrand?

One approach is to use cylindrical shells of thickness [itex]\Delta x[/itex]. You will need to find the equation of the circle. Its center is at (R, 0) and its radius is r. Find an expression for the incremental area [itex]\Delta A[/itex]. I'm sure your text has an explanation of the shell method and several examples.
oops, forgot the dx.

We were doing examples in class where we would find the distance from the outer function to the origin and from the inner function to the origin(where we set the problem at).

Getting the thickness of the torus could be done by rotating some [tex]\Deltax[/tex] about the x axis(to get the area of the circle and then getting the actual shape of the torus would require you to rotate the area about x=0 right?
Mark44
#7
Jul13-10, 09:54 PM
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P: 21,283
Can you edit your previous post - the part with [ tex]Deltax[ /tex]? By putting a space between Delta and x. What you have is causing my browser to render a big empty box that's very wide.
Quote Quote by nlsherrill View Post
oops, forgot the dx.

We were doing examples in class where we would find the distance from the outer function to the origin and from the inner function to the origin(where we set the problem at).

Getting the thickness of the torus could be done by rotating some [tex]\Delta x[/tex] about the x axis
??? I don't get what you're saying here.
Quote Quote by nlsherrill View Post
(to get the area of the circle and then getting the actual shape of the torus would require you to rotate the area about x=0 right?
I suggested using cylindrical shells. Stewart seems to be using disks of thickness [itex]\Delta x[/itex]
Mark44
#8
Jul13-10, 10:04 PM
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P: 21,283
Quote Quote by Gregg View Post
Isn't the volume of a torus [itex] 2\pi^2r^2 R [/itex]?
Yes, and is the answer in the back of the book.
nlsherrill
#9
Jul14-10, 07:17 AM
P: 322
Okay so I got an annoyingly close answer according to the back of the book.

My answer is 4[tex]\PiR[/tex][tex]\int\sqrt{r^2-y^2}dy[/tex]

^^ I don't know why 4PiR isnt showing up before the integral. Latex is still new to me and I have no idea why my posts writing is being split so far apart either.

The answer is has an 8 instead of a 4 though. Would this have anything to do with the limits of integration being from 0->R instead of maybe -R->R?

Thanks for help thus far.
Mark44
#10
Jul14-10, 09:36 AM
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P: 21,283
Quote Quote by nlsherrill View Post
Okay so I got an annoyingly close answer according to the back of the book.

My answer is 4[tex]\pi R[/tex][tex]\int\sqrt{r^2-y^2}dy[/tex]

^^ I don't know why 4PiR isnt showing up before the integral. Latex is still new to me and I have no idea why my posts writing is being split so far apart either.
The LaTeX parser doesn't know what \PiR is, just as it doesn't know what \Deltax is. Please edit both your posts by putting a space between \pi and R and between \Delta and x.

Also, rather than having multiple pairs of tex brackets, it works better to put the entire expression inside one pair of tex brackets, like this:
[tex]4\pi R \int\sqrt{r^2-y^2}dy[/tex]

Click on the expression above to see what I did.

Quote Quote by nlsherrill View Post

The answer is has an 8 instead of a 4 though. Would this have anything to do with the limits of integration being from 0->R instead of maybe -R->R?
You are using "washers" of thickness [itex]\Delta y[/itex]. What is the range of y values from which [itex]\Delta y[/itex] is a subinterval? It is not 0 to R and it is not -R to +R.


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