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About LinkBacksFind the surface area given by rotating the following curve about the y-axis.
y=
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
Find the surface area given by rotating the following curve about the y-axis.
y=
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
LaTeX hasn't been working for about a week. Hadn't you noticed?
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}}
Your first problem is that
squared is
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
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.
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