
#1
Jul1310, 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 donutshaped 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(Rr)^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 outerinner functions and integrated. Heres a picture from the book that may help.. [/b] 



#2
Jul1310, 08:36 PM

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P: 20,943

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. 



#3
Jul1310, 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]?




#4
Jul1310, 09:19 PM

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P: 20,943

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. 



#5
Jul1310, 09:27 PM

P: 463

Isn't the volume of a torus [itex] 2\pi^2r^2 R [/itex]?




#6
Jul1310, 09:29 PM

P: 322

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? 



#7
Jul1310, 09:54 PM

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P: 20,943

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.




#8
Jul1310, 10:04 PM

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P: 20,943





#9
Jul1410, 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^2y^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. 



#10
Jul1410, 09:36 AM

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P: 20,943

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^2y^2}dy[/tex] Click on the expression above to see what I did. 


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