And that integral is the result from integrating the volume of the "upper half" of the torus, because the square root will only provide a positive height for the cylindrical shell (except that 'Y' should also be an 'x'). The radius of the shell is 3-x , the height above the x-axis is \sqrt{4-x^2}, so the complete volume integral is
V = 2 \int_{-2}^{2} 2\pi (3-x) \sqrt{4-x^2} dx
(your choice of whether to attach the '2' to the square root or put it out in front to get the full height of the shells), which becomes
V = 4\pi \int_{-2}^{2} (3-x) \sqrt{4-x^2} dx .
The integral will have two terms, but since the function x \sqrt{4-x^2} has odd symmetry and we are integrating symmetrically about the y-axis, the second integral term is zero, reducing our labor to
V = 12\pi \int_{-2}^{2} \sqrt{4-x^2}dx ,
which, after a trig substitution, gives us V = 12\pi \cdot 2\pi = 24\pi^{2} .
The washer method looks worse to set up, but the integral reduces to something similar. We would use x =\pm \sqrt{4-y^2} and we want both signs. The inner radius of the washer will be 3 minus that square root, the outer radius, 3 plus that radical, and -- again integrating only the "upper half" of the torus -- the volume is
V = 2 \int_{0}^{2} \pi [(3 + \sqrt{4-y^2})^2 - (3 - \sqrt{4-y^2})^2 ] dy ,
which algebraically simplifies to
V = 2\pi \int_{0}^{2} 12\sqrt{4-y^2} dy = 24\pi \cdot \pi
I went and looked it up and now I understand why you asked me why I was talking about surface area. My apologies: I find that I should have said "Pappus' Second Theorem" or "Pappus' Centroid Theorem". (A lot of calculus books I've come into contact with never mention the First Theorem, so I forgot about that one...) The Centroid Theorem says the volume of the solid of revolution equals the area of the revolved closed curve times the circumference of the path covered by the curve's centroid. We have a circle of radius 2 and center at the origin being carried around the line x = 3, so the volume is just
V = (\pi \cdot 2^2) \cdot (2\pi \cdot 3) = 24\pi^{2}
