Surface of revolution using cylindrical polars Consider the surface of revolution of the curve $$y = x^2$$ where $0 < x < 1$. By writing a suitable integral, show that the area of this surface is 3.81 units. (You
are advised to work in cylindrical polar). 
 A: This integral is very simple in cartesian coordinates.  In this case, an element of surface area is $2 \pi x ds$, where $ds$ is an element of arc length along the given curve.  Therefore
$$S = 2 \pi \int_0^1 ds \, x = 2 \pi \int_0^1 dx \, x \sqrt{1+\left ( \frac{dy}{dx}\right)^2} = 2 \pi \int_0^1 dx \, x \sqrt{1+4 x^2}$$
which evaluates very easily  to
$$S = \pi \int_0^1 du  \,  \sqrt{1+4 u} = \frac{\pi}{4} \frac{2}{3} \left[(1+4 u)^{3/2}\right]_0^1 = \frac{\pi}{6} \left ( \sqrt{125}-1\right)$$
which, by the way, is about 5.33 units, not 3.81 units.
A: Employ an understanding of surface of revolution of a curve! :P
That is, we make the polar coordinate substitutions first: $x=r\cos \theta$ and $y=r\sin \theta$. (Just recall the unit circle.) From these substitutions we arrive $r\sin \theta=r^2\cos^2\theta$, which reduces to $r=\frac{\sin \theta}{\cos^2\theta}$.
From here, we now have that the area is $S=\int_0^{1} yds$ since we are rotating about the $x$-axis. Note that $ds=\sqrt{r^2+r'^2}d\theta$. Want to take it from here?
