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volume of a torus, in terms of big radius R and little radius r, washer method

skatenerd

Active member
Oct 3, 2012
114
This problem is really making my head spin. I'm in university calculus II and my teacher loves giving us extra homework that is supposed to really challenge us. This is problem 1 of 5.
"Let 0<r<R and x2+(y-R)2=r2 be the circle centered at (0,R) of radius r. Revolving the disk enclosed by that circle about the x-axis generates a torus. Using the washer method obtain the volume of that torus."

So just a disclaimer, I have NO intention of cheating on this. I'm in the class to learn. I have gotten so far on this problem and now I'm stuck which is very frustrating since figuring out everything before now took a while. If anybody could just let me know if I did something wrong or not that would be awesome.

So far, I made a graph, and made an infinitesimally small slice through the torus, which I needed to find the area of. To do this I knew I needed the equation solved for y, so I did that and got:
y=R+(root(r2-x2)) for the top half of the circle and y=R-(root(r2-x2)) for the bottom half. I used the top half as the outer radius of the washer and the bottom half as the inner radius. From this I got an integral:
pi ( int (R+(root(r2-x2)))^2-(R-(root(r2-x2)))^2 ) dx
this ends up being:
4piR ( int (root(r2-x2)) ) dx

and from there I tried using a couple different u-substitutions and nothing worked and I got frustrated, then came here. Any help would be appreciated! Thanks in advance.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I'm with you up to here:

$\displaystyle V=4R\pi\int_{-r}^r \sqrt{r^2-x^2}\,dx$

I would then use symmetry (integrand is even and limits are symmetric about x = 0) to write:

$\displaystyle V=8R\pi\int_{0}^r \sqrt{r^2-x^2}\,dx$

Now, try the trigonometric substitution:

$\displaystyle x=r\sin(\theta)\,\therefore\,dx=r\cos(\theta)\,d \theta$
 

skatenerd

Active member
Oct 3, 2012
114
Alright so I tried that trig substitution.
The r^2 was easily factored out of the square root, and i ended up with:
8piR ( int r(root(1-sin^2(theta))) ) from 0 to r.
This then equals
8piR ( int rcos(theta) ) from 0 to r.
this is also known as
8piR(rsin(theta)). This part of the trig substitution is always trickiest for me. However since we came out with rsin(theta) which was the substitution in the first place, I feel like I cant just make that rsin(theta) = x. (correct me if I'm wrong, I feel like I usually am at that part).
so now I have 8piR(x) from 0 to r. Solve this out ends up giving me 8piRr which is not the volume of a torus...I'm looking for the answer 2pi^2r^2R, right?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, that is what you are looking to get.

You have made two errors with the substitution:

i) You did not substitute correctly for the differential dx.

ii) You did not change your limits of integration in accordance with the substitution.

The original limits of integration are in terms of x and after the substitution you want them to be in terms of θ.

So, take the substitution, solve for θ, then plug in the original limits for x to get the new limits.
 

skatenerd

Active member
Oct 3, 2012
114
Okay thanks. I saw what I was doing wrong, careless little mistake I guess. So I finished the problem, and came up with the right answer. One more question though, just to make sure I would be able to do this on a test if need be.
During the end of the trig substitution when plugging back in to get rid of the thetas, I get confused. I understand that 1/2(theta) would just be 1/2(arcsinx), but in the other part of the equation you have 1/4(sin(2(theta))). Trying to plug back in to get it in terms of x here is really confusing for me. The only reason I knew what it was to get the right answer for my problem was because this was covered in class once, but I never understood what the teacher did in class to answer it. If you could try explaining why 1/4(sin(2(theta))) would end up being 1/2(root(r2-x2)) that would be awesome. I know that the imaginary triangle has sides of x for opposite, r for hypotenuse and root(r2-x2) for adjacent, but I just can't understand how to use that for 1/4(sin(2(theta))).
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
This is how I would handle it. We have:

$\displaystyle V=8\pi R\int_0^r\sqrt{r^2-x^2}\,dx$

Using the substitution:

$\displaystyle x=r\sin(\theta)\,\therefore\,dx=r\cos(\theta)\,d \theta$

We have:

$\displaystyle V=8\pi R\int_{\theta(0)}^{\theta(r)}\sqrt{r^2-r^2\sin^2(\theta)}\,r\cos(\theta)\,d\theta$

Now, to change the limits of integration, observe we have:

$\displaystyle \theta(x)=\sin^{-1}\left(\frac{x}{r} \right)$ and so:

$\displaystyle \theta(0)=\sin^{-1}\left(\frac{0}{r} \right)=0$

$\displaystyle \theta(r)=\sin^{-1}\left(\frac{r}{r} \right)=\frac{\pi}{2}$

Now, on the interval $\displaystyle \left(0,\frac{\pi}{2} \right)$, we have the sine and cosine functions being non-negative, hence we may write the integral as:

$\displaystyle V=8\pi R\int_{0}^{\frac{\pi}{2}}r\sqrt{1-\sin^2(\theta)}\,r\cos(\theta)\,d\theta$

$\displaystyle V=8\pi Rr^2\int_{0}^{\frac{\pi}{2}}cos^2(\theta)\,d\theta$

Now, using the identity $\displaystyle cos^2(x)=\frac{1+\cos(2x)}{2}$ we have:

$\displaystyle V=4\pi Rr^2\int_{0}^{\frac{\pi}{2}}1+\cos(2\theta)\,d \theta$

Hence:

$\displaystyle V=4\pi Rr^2\left[\theta+\frac{1}{2}\sin(2\theta) \right]_{0}^{\frac{\pi}{2}}=4\pi Rr^2\left(\frac{\pi}{2} \right)=2\pi^2Rr^2$
 

skatenerd

Active member
Oct 3, 2012
114
No see I understand all of that. What I want to clear up and figure out is the last part of that post you just left. When you substitute back in to get rid of the thetas. How do you got from 1/2(sin2theta) to x(root(r2-x2​)) ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I don't follow. Once we have made the substitution, we are no longer concerned with x. Everything is in terms of the new variable θ.
 

skatenerd

Active member
Oct 3, 2012
114
I see the confusion. I guess it is kinda weird that when I do a substitution with a definite integral I don't usually change the limits of integration because I don't plug them in until my integral is completely solved. At that point it wouldn't really matter if the bounds were changed accordingly to the substitution because you have the answer to the integral which is what it would have been anyways. But I see how the way you do it would make the trig substitution a good deal easier in the end, so I'll try to make a habit out of doing it that way. Thanks.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
It's really a matter of preference; I just find it simpler for a definite integral to change over to the new variable completely, and then leave the old variable behind.