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