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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).

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You are advised not to direct imperatives at us and to formulate questions instead. – joriki Dec 2 '12 at 16:28
A surface of revolution lives in $3$-space with coordinates $x$, $y$, $z$. What is the axis of revolution in your case? What is the plane containing the curve $y=x^2$ with respect to the axis of revolution? – Christian Blatter Dec 2 '12 at 21:04

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.

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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?

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Can you show some working please, I have attempted this but I get a really long and complex integral and finding it hard to solve this problem – antinode Dec 5 '12 at 9:56
@antinode I am also finding it hard to solve this integral. I'm pursuing alternative methods. – 000 Dec 5 '12 at 23:18

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