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 2∏ then, 2∏ thus, ∏
you're right ... I had area of a circle on the brain instead of circumference. corrected.
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 2∏ 2∏ 2∏ 2∏ 2∏ ∏/2 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?
you're done.
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