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).
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
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? |
|||
