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The given curve is rotated about the $y$-axis. Find the area of the resulting surface.

$y= (1/4 x^2) - (1/2 \ln x)$. $x$ is in between 1 and 2 (including 1 and 2). If anyone could please point me in the right direction of how to solve this I would be very grateful :)

Daniel Fischer
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    The relevant formula is shown in http://math.stackexchange.com/questions/6979/areas-versus-volumes-of-revolution and probably in your book. Have you tried anything? – Ross Millikan Mar 09 '11 at 18:36
  • @Ross: Not quite: That formula is for rotation of the graph around the $x$-axis... – Arturo Magidin Mar 09 '11 at 19:10
  • @Arturo: true, but I hoped OP could fix that. Should I delete the comment? – Ross Millikan Mar 09 '11 at 19:15
  • @Ross: Well, now that we've added a small note that there is a need to amend the formula, no, you shouldn't. (At least, until the OP starts asking questions). – Arturo Magidin Mar 09 '11 at 19:16
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    @user8032: Is your relation between $y$ and $x$ really $y=\frac{1}{4} x^2 - \frac{1}{2} x \ln x$? – Fabian Mar 09 '11 at 19:25

2 Answers2

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The formula for the area of a surface of revolution about the $y$-axis formed by $(x(t),y(t))$ on $a\le t\le b$ is $$2\pi \int_a^b x(t)\sqrt{(x'(t))^2+(y'(t))^2} dt.$$ In your case, let $x(t)=t$ so that $y(t)=\frac{1}{4}t^2-\frac{1}{2}t\ln t$.

Isaac
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  • $a=1$, $b=2$, $f(x)=\frac{1}{4} x^2 - \frac{1}{2} x \ln x$, $f'(x)= \frac{d}{dx} f(x)$, ... – Fabian Mar 09 '11 at 19:04
  • Something does not look right. This is not an integral which would appear in a homework... – Fabian Mar 09 '11 at 19:06
  • @Arturo Magidin: ok thanks, I was just trying to get some of the unknowns of the OP and then figured out that I couldn't solve the integral ;-) – Fabian Mar 09 '11 at 19:11
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    @Arturo: I think I've got the formula right this time, but it's still not a nice integral... – Isaac Mar 09 '11 at 19:18
  • We haven't covered this topic in class yet and I'm only familiar with with revolutions about the x-axis. I tried playing around with the provided formula but seem to keep going in circles –  Mar 10 '11 at 14:38
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The integral, though not nice is do-able. Generally for a function y=f(x), the formula for the surface area, as suggested by Isaac, is:

$$\sigma=2\pi\int^b_a{f(x)\sqrt{1+[f'(x)]^2} dx}$$ (where $\sigma$ is surface area)

So You result in

$$\sigma=2\pi\int^2_1{\left(\frac{x^2}{4}-\frac{ln(x)}{2}\right)\sqrt{1+\left[\frac{x}{2}-\frac{1}{2x}\right]^2}dx}$$

You will find that this integral simplifies to:

$$\sigma=\frac{\pi}{4}\int^2_1{\left[x^2-2ln(x)\right]\left(\frac{x^2+1}{x}\right) dx}$$

which is: $$\sigma=\frac{\pi}{4}\left[\left.\left(\frac{x^4}{4}+\frac{x^2}{2}\right)\right|^2_1-2\left.\left(\frac{x^2}{2}\left(ln(x)-\frac{1}{2}\right)+\frac{ln^2(x)}{2}\right)\right|^2_1\right]$$ $$\sigma=\frac{\pi}{4}\left[\left(\frac{21}{4}\right)-2\left(\frac{ln^2(2)}{2}+2ln(2)-\frac{3}{4}\right)\right]$$ $$\sigma=\left[\frac{21\pi}{16}-\frac{\pi}{2}\left(\frac{ln^2(2)}{2}+2ln(2)-\frac{3}{4}\right)\right]$$ $$\sigma=\left[\frac{21\pi}{16}-\frac{\pi ln^2(2)}{4}-\pi ln(2)+\frac{3\pi}{8}\right]$$ $$\sigma=\left[\frac{21\pi}{16}+\frac{3\pi}{8}-\frac{\pi ln^2(2)}{4}-\pi ln(2)\right]$$ $$\sigma=\left[\frac{27\pi}{16}-\frac{\pi ln^2(2)}{4}-\pi ln(2)\right]$$ $$\sigma=\left[\frac{27\pi}{16}-\pi ln(2) \left(ln(e\cdot\sqrt[4]{2}) \right) \right]$$

This is the most simplifies exact version obtainable which is approximately:

$$\sigma\hspace1ex\dot{=}\hspace1ex\left(5.30144-2.55493\right)$$ $$\sigma\hspace1ex\dot{=}\hspace1ex 2.74651$$

And so to conclude,

$\therefore$ the surface area of $\left[y=f(x)=\left(\frac{x^2}{4}\right)-\left(\frac{ln(x)}{2}\right)\right]$ revolved 360 degrees about the x-axis from $\left.\right|^2_{x=1}$ is about $2.74651units^2$

Keith Afas
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