Surface of revolution using cylindrical polars

Question

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).

Answer 1

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?