y=1/6(x³) + 1/(2x) , 1 ≤ x ≤ 2
I don't really understand how to solve this...
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arc length ...
surface area ...
where is the radius of revolution as a function of
ok for the arc length i have
∫√[1- [(x⁴-1)/(2x²)]²] dx from 1 to 2
∫√[1- [(x⁸-2x⁴+2)/(4x⁴)] dx from 1 to 2
i can't get past that.
For the srface area i have
2∏∫ [(x³)/6 + 1/(2x)] √[1- [(x⁸-2x⁴+2)/(4x⁴)] dx
and I can't get past that either.
Originally Posted by hlpplz y=1/6(x³) + 1/(2x) , 1 ≤ x ≤ 2
can you finish ?
how did you get
i thought the equation for surface area was
2∏∫ f(x) √[1+f'(x)] dx
wouldn't that make it be
you're right ... I had area of a circle on the brain instead of circumference.
so the i have
am i right so far?
you can only multiply one of the two factors in the integrand by 2, not both.
ok from this
what i did was just multiply it all out and got
is that better?
integrate at this point. don't mess with the algebra anymore ... too big a chance of making a bonehead error.
ok so for the final result i was able to get
is that correct?
Originally Posted by hlpplz ok so for the final result i was able to get
is that correct? my calculator says
and for the arc length i ended up getting
is that right?
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