# Surface Area of a Revolution

• Apr 30th 2011, 09:24 AM
dbakeg00
Surface Area of a Revolution
Find the surface area given by rotating the following curve about the y-axis.

y= $\frac{x^4}{4} + \frac{1}{8x^2}$ $0\leq x \leq 3$

$\frac{dy}{dx}=x^3-\frac{1}{4x^3}$

$(\frac{dy}{dx})^2=x^6-\frac{1}{2x^6}-\frac{1}{2}$

$s=\int^3_0 2\pi*x*\sqrt{1+x^6-\frac{1}{2x^6}-\frac{1}{2}}$

$s=\int^3_0 2\pi*x*\sqrt{x^6-\frac{1}{2x^6}+\frac{1}{2}}$

That is as far as I can get. Can someone tell me how to take this integral? I need to take it just using algebraic manipulation and u-substitution if necessary. Thanks
• Apr 30th 2011, 09:26 AM
dbakeg00
Whats going on here? I couldn't get the preview function to work but I thought it was just broken or something. I'm using LaTex the way I always have...am I doing it wrong now? Did something change?
• Apr 30th 2011, 10:18 AM
HallsofIvy
A number of new "buttons" apparently connected to LaTeX that don't showing up properly seem to been added. I suspect that they tried to add something to make it easier to write LaTeX and blew up in the forum's face!

You appear to have written
y=\frac{x^4}{4} + \frac{1}{8x^2} 0\leq x \leq 3

\frac{dy}{dx}=x^3-\frac{1}{4x^3}

(\frac{dy}{dx})^2=x^6-\frac{1}{2x^6}-\frac{1}{2}

s=\int^3_0 2\pi*x*\sqrt{1+x^6-\frac{1}{2x^6}-\frac{1}{2}}

s=\int^3_0 2\pi*x*\sqrt{x^6-\frac{1}{2x^6}+\frac{1}{2}}

$\frac{1}{4x^3}$
squared is
$\frac{1}{16x^2}$
not
[tex]\frac{1}{2x^6}

Now, for the rest-
That's a very common trick for this kind of problem- just because it is so hard to find f so that
$\sqrt{1- \left(\frac{df}{dx}\right)^2}$
is easy to integrate.
Notice that the only difference between your last line and your third is that the last line has
\sqrt{1+ \frac{1}{2}+ \frac{1}{16x^2}}
while the third line has
\sqrt{1+ \frac{1}{2}- \frac{1}{16x^2}}
is that the middle term has changed from "-" to "+".

That is, it is just the difference between (a- b)^2= a^2- 2ab+ b^2 and a^2+ 2ab+ b^2= (a+ b)^2.