Area between rotated figure Find the area of the zone of a sphere formed by revolving the graph of 
$y = \sqrt{r^2 − x^2}$ , $0 \leq x \leq a$, about the $y$-axis. Assume that $a < r$.
 A: Draw a picture. The surface of the region we obtain consists of (i) the upper curvy component that is part of the full surface of the sphere and (ii) a middle component which is the curvy surface of a cylinder and (iii) the lower curvy component. The lower component has the same area as the upper one. 
But I would interpret "zone of the sphere" as not including the cylindrical part. So all we need to do is to find the surface area of the upper component, and double the result.  
By a formula that I hope is familiar, the area (i) is equal to
$$\int_0^a 2\pi x \,ds,$$
where 
$$ds=\left(1+\left(\frac{dy}{dx}\right)^2\right)^{1/2}\,dx.$$
We have $y=(r^2-x^2)^{1/2}$. When we do the differentiation, we get $\dfrac{dy}{dx}=-\dfrac{x}{(r^2-x^2)^{1/2}}$. Square, add $1$, take the square root. There is a nice amount of simplification, and we end up with $\dfrac{r}{(r^2-x^2)^{1/2}}$.
Thus the surface area (i) is equal to
$$\int_0^a 2\pi r \frac{x}{(r^2-x^2)^{1/2}}\,dx.$$
This integral yields to the substitution $u=r^2-x^2$. Indeed, we already know the antiderivative, since $-\dfrac{x}{(r^2-x^2)^{1/2}}$ was what we obtained when differentiating $(r^2-x^2)^{1/2}$. 
