Find the SA of a torus I have been trying to find the surface area of the torus generated by the rotation of $(x-R)^2 + y^2 = r^2$ about the y axis. I tried to use the equation:
$$\int_a^b2\pi y\sqrt{1+\left(\frac {dy}{dx}\right)^2}dx$$
I know that the derivative of the equation is 
$$\frac {dy}{dx}= \frac {-(x+R)}{\sqrt{r^2-(x+R)^2}}$$
If I am correct, the limits of integration are $R+r$ and $R-r$. When I plug it in the equation I get:
$$2\pi\int_{R-r}^{R+r} \sqrt{r^2+(x+R)^2} * \sqrt{1+\frac {(x+R)^2}{r^2-(x+R)^2}}dx $$
When you simplify, the first square root cancels out and in the second square root we are left with r^2:
$$2\pi\int_{R-r}^{R+r}rdx$$
Clearly, this isn't going to lead to the right answer of $4\pi^2Rr$. What am I doing wrong?
 A: The formula you quoted is for rotating about the $x$-axis. To modify it for rotation about the $y$-axis, interchange the roles of $x$ and $y$. So the surface area, because of symmetry, is
$$2\int_{y=0}^r 2\pi x\sqrt{1+\left(\frac{dx}{dy}\right)^2}  \,dy.$$
We have $x-R=\pm\sqrt{r^2-y^2}$ (the "inner" part of the circle is $x=R-\sqrt{r^2-y^2}$ and the "outer" part of the circle is $x=R+\sqrt{r^2-y^2}$).
Differentiate. We get $\frac{dx}{dy}=\mp\frac{y}{\sqrt{r^2-y^2}}$. Square, add $1$, take the square root. We get after some simplification that the area is
$$2\int_{y=0}^r 2\pi\left(R-\sqrt{r^2-y^2}\right)\frac{r}{\sqrt{r^2-y^2}}\,dy+2\int_{y=0}^r 2\pi\left(R+\sqrt{r^2-y^2}\right)\frac{r}{\sqrt{r^2-y^2}}\,dy.$$
(The first integral gives the inner surface area, the second gives the outer surface area.)
The two integrals can be profitably combined, to give
$$2\int_{y=0}^r 2\pi\frac{2Rr}{\sqrt{r^2-y^2}}\,dy.$$
Now the integration is straightforward. We get $(2)(2\pi)(2Rr)(\pi/2)$.
Remark: At the beginning, to keep things fully familiar, it might be more comfortable to take the circle $x^2+(y-R)^2=r^2$, and rotate it about the $x$-axis. Then the surface of the torus is the surface generated by the upper half of the circle, that is, $y=R+\sqrt{r^2-x^2}$, plus the area generated by the lower half of the circle. Same calculation, really, but it may seem easier.
