Find the area of the surface obtained The curve , x^2 , is rotated about the y-axis.
(a) Find the area of the resulting surface.
(b) Find the area of the surface obtained by rotating the curve
in part (a) about the x-axis.

Okay Part A was easy for me. I just found dy.dx and used the ds formula and put ds in the area formula.
But for part b, it asks the same thing except it wants to rotate about the x-axis. How would I do this? 
Thank you,
 A: What we rotate was incompletely specified. Say it is the part of the curve $y=x^2$ from $x=0$ to $x=c$, where $c\gt 0$. 
To solve the second problem,  interchange the roles of $x$ and $y$. So our curve is $x=\sqrt{y}$. We have
$$\frac{dx}{dy}=\frac{1}{2\sqrt{y}},$$
so we end up with the integral
$$\int_0^{\sqrt{c}} 2\pi y \sqrt{1+\frac{1}{4y}}\,dy.$$
So we want to integrate $\pi \sqrt{4y^2+y}$. Somewhat unpleasant, but doable by completing the square, and making the appropriate substitution, either a secant or a hyperbolic cosine. 
Remark: Slightly more mechanically, we can observe that the surface area (of the curved part) is
$$\int_0^c 2\pi x^2 \sqrt{1+4x^2}\,dx.$$
We get an integral that is not very pleasant. But it yields (after a while) to the substitution $2x=\tan t$ or $2x=\sinh t$. 
A: To rotate a function around an axis you parametrize it correctly, in this case the parametrization is:
$\sigma(\phi,t)=(t,t^2\cos\phi,t^2\sin\phi)$ Where the domain of $t$ is the original domain of $x$, and $0\le\phi\le 2\pi$.
Effectively what happens in this parametrization, is $(t^2\cos\phi,t^2\sin\phi)$ represents a "circle" where the radius varies exactly like the function we wish to rotate. Then the "$(t)$" part is just the $x$ axis, which changes linearly since we are rotating the function around the $x$ axis.
