volumes by cylindrical shells problem. Find the volume of the solid obtained by rotating about the x-axis the region enclosed by the curve
$$x^2+(y-R)^2=r^2 (R>r>0) $$
 A: Draw a picture. The region you are rotating is a "circle" (disk) of radius $r$ with centre at $(0,R)$. The solid we obtain is a torus (doughnut). 
Take a thin strip at height $y$, of thickness "$dy$". We will be rotating such strips about the $x$-axis, and "adding up" (integrating).
To avoid minus signs, let's rotate the part of the strip that is in the first quadrant, and double the answer at the end. 
The width of the strip is then $x$. So the half-torus has volume
$$\int_{y=R-r}^{R+r} 2\pi xy \,dy.$$
To do the integration, we need to express $x$ in terms of $y$. From $x^2+(y-R)^2=r^2$ we get $x=\sqrt{r^2-(y-R)^2}$.
We leave the integration to you, but strongly suggest you make the substitution $u=y-R$.  
Remark: If you are accustomed to using the shell method for rotation about the $y$-axis, and don't like change, you can interchange the roles of $x$ and $y$, rotating the region inside $(x-R)^2+y^2=r^2$ about the $y$-axis. 
A: Hint: 
$x^2+(y-R)^2=r^2$ . Do you see that the curve is a circle.? 
The center lies on $(0,R)$. And you get a circle with radius $r$. What solid do you get when you rotate a circle along the axis? Its looks like a circular cylinder disc.
It looks something like this(Its called Torus):

Volume of the given Torus would be: Volume of cylinder= Cross-sectional area $\times$ Length=$\pi r^2 \cdot 2 \pi R=2 \pi^2Rr^2$ 
A: Note that the curve (a circle) doesn't intersect the $x$-axis. (Why not?) Moreover, it is symmetric about the $y$-axis, so the volume of its solid of revolution will be twice the volume of the solid of revolution of the region bounded on the left by the $y$-axis and on the right by the given curve. Thus, by the cylindrical shells method, the desired volume will be $$2\cdot 2\pi\int_{R-r}^{R+r}y\sqrt{r^2-(y-R)^2}\,dy.$$ Can you take it from there (perhaps by making a convenient $u$-substitution)?
