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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 :)

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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?2011-03-09
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    @Ross: Not quite: That formula is for rotation of the graph around the $x$-axis...2011-03-09
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    @Arturo: true, but I hoped OP could fix that. Should I delete the comment?2011-03-09
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    @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).2011-03-09
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    @user8032: Is your relation between $y$ and $x$ really $y=\frac{1}{4} x^2 - \frac{1}{2} x \ln x$?2011-03-09

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