# Thread: Shell integration of a donut

1. ## Shell integration of a donut

Question: The region inside the circle y^2 + (x-2)^2 = 1 is revolved about the y-axis to form a donut. Use the method of shells to find the volume of the donut.

Ok soo... i tried shell integration and when i subbed in my limits (y=1, y=-1), i end up getting zero.

$\displaystyle V= \int 2*pi*([sqrt(1-y^2)] - 2)*y dy from -1 to 1$

I distributed that last y to the sqrt and 2.
Then I did a substitution (1-y^2) = t which changed the boundaries to t=0 and t=0... so that gives zero.

And then I integrated the second term (2y) which comes out to y^2 from y=-1 to y=1 also gives zero...

I tried integrating with respect to x (from x=1 to x=3), but that was horrible and also comes out to zero...

I don't really know where I'm going wrong (probably the boundaries) but any help would be appreciated

2. Since this is rotated around the y axis, using shells, you should be integrating with respect to x, not y.

From $\displaystyle (x-2)^2+ y^2= 1$, $\displaystyle y= \pm\sqrt{1- (x-2)^2}= \pm\sqrt{-5+ 4x- x^2}$ with x between 2-1= 1 and 2+1= 3. Each shell is a cylinder with height $\displaystyle 2\sqrt{-5+ 4x- x^2}$ and radius x so the volume is given by 2\pi\int_{x=1}^3 x\sqrt{-5+ 4x- x^2}dxp[/tex]

I think I would be inclined, rather, to use "disks" or "washers". From $\displaystyle (x-2)^2+ y^2= 1$, $\displaystyle x= 2\pm\sqrt{1- y^2}$ for each y. That is, a cross section of the solid at each y is a "washer" lying between the two circles of radius $\displaystyle 2- \sqrt{1- y^2}$ and $\displaystyle 2+ \sqrt{1- y^2}$. You can calculate the area of the "washer" between the two circles as the difference between the area of the two circles: $\displaystyle \pi\left(2+ \sqrt{1- y^2}\right)^2- \left(2- \sqrt{1- y^2}\right)^2= 8\pi\sqrt{1- y^2}$ and the volume is $\displaystyle 8\pi\int_{y= -1}^1 \sqrt{1- y^2}dy$

3. Thanks for the response

I tried integrating again with respect to x,

$\displaystyle 2\pi\int_{x=1}^3 x\sqrt{-3+ 4x- x^2}dx$

the only difference is that its -3 in the square root and not -5.

More importantly, after integrating this, you get

$\displaystyle 2\pi[1/3*\sqrt{1-(x-2)^2}*(x^2-x-3)-arcsin(2-x)]$ from x=1 to x=3

If you sub in x=3 and x=1 you actually end up with zero XD

So I went over my steps and noticed that at one point there is a spot of discontinuity... well maybe, im not sure.

$\displaystyle 2\pi\int_{x=1}^3 x\sqrt{1-(x-2)^2}dx$

Substitute s = x-2;
ds=dx

$\displaystyle 2\pi\int_{s=-1}^1 (s+2)\sqrt{1-s^2}dx$

Substitute t=arcsin(s)
$\displaystyle dt=1/\sqrt{1-s^2}ds$

This is where the discontinuity would be... if s=+-1, then its undefined.
But I don't know what this changes?? This might not even be the problem...

I checked my integration with an online integrator (wolfram mathematica) and it seems to be poppin out the same answer so I got no idea where I'm going wrong >.<

4. Hey laughingman, when I tried your integral of:

2pi [(1/3)(sqrt(1-(x-2)^2))(x^2-x-3) - arcsin(2-x)] from x=1 to x=3, I ended up getting pi inside the square brackets which resulted in the answer being 3pi with the 2pi on the outside, so maybe just trying subbing x=1 and x=3 again.

But my question to you is, how did you integrate:

2pi[ int(x(sqrt (1-(x-2)^2))) ] from x=1 to x=3 into the integral you got above? I'm curious because I couldn't do it myself