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Math Help - Arc Length and Area Surface by revolution

  1. #1
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    Arc Length and Area Surface by revolution

    y=1/6(x) + 1/(2x) , 1 ≤ x ≤ 2

    I don't really understand how to solve this...
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  2. #2
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    arc length ...

     <br />
S = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx<br />

    surface area ...

     <br />
A = \int_a^b r(x) \cdot \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx<br />
    where r(x) is the radius of revolution as a function of x
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  3. #3
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    ok for the arc length i have

    ∫√[1- [(x⁴-1)/(2x)]] 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∏∫ [(x)/6 + 1/(2x)] √[1- [(x⁸-2x⁴+2)/(4x⁴)] dx

    and I can't get past that either.
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  4. #4
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    Quote Originally Posted by hlpplz View Post
    y=1/6(x) + 1/(2x) , 1 ≤ x ≤ 2
    \frac{dy}{dx} = \frac{x^2}{2} - \frac{1}{2x^2}

    \left(\frac{dy}{dx}\right)^2 = \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}

    1 + \left(\frac{dy}{dx}\right)^2 = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2

    \sqrt{\left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2} = <br />
\frac{x^2}{2} + \frac{1}{2x^2}

    S = \int_1^2 \frac{x^2}{2} + \frac{1}{2x^2} \, dx

    A = \pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)^2\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx<br />

    can you finish ?
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  5. #5
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    how did you get

    <br /> <br />
A = \pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)^2\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx<br />

    i thought the equation for surface area was

    2∏∫ f(x) √[1+f'(x)] dx

    wouldn't that make it be

    2∏ <br /> <br />
\int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx<br />

    then,

    2∏ <br /> <br />
\int_1^2 \frac{1}{2}\left(\frac{x^3}{3} + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) \, dx<br />

    thus,

    <br /> <br />
\int_1^2 \left(\frac{x^3}{3} + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) \, dx<br />
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  6. #6
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    you're right ... I had area of a circle on the brain instead of circumference.

    corrected.

    <br />
A = 2\pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx<br />
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  7. #7
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    so the i have

    <br /> <br />
\int_1^2 \left(\frac{x^3}{3} + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) \, dx<br />

    <br /> <br />
\int_1^2 \left(\frac{x^5}{3} + \frac{x^3}{3x^2} + \frac{x^2}{x} + \frac{x}{x^3}\right) \, dx<br />

    <br /> <br />
\int_1^2 \left(\frac{x^5}{3} + \frac{x}{3} + {x} + \frac{1}{x^3}\right) \, dx<br />

    <br /> <br />
\int_1^2 \left(\frac{x^5}{3} + \frac{4x}{3} + \frac{1}{x^3}\right) \, dx<br />

    <br /> <br />
\left(\int_1^2 \frac{x^5}{3} + \int_1^2 \frac{4x}{3} + \int_1^2 \frac{1}{x^3} \, dx \right)<br />

    am i right so far?
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  8. #8
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    you can only multiply one of the two factors in the integrand by 2, not both.

    \pi \int_1^2 \left(\frac{x^3}{3} + \frac{1}{x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx<br />
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  9. #9
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    ok from this

    <br /> <br />
\int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx<br />

    what i did was just multiply it all out and got

    2∏ <br /> <br />
\int_1^2 \left(\frac{x^5}{12} + \frac{x^3}{12x^2} + \frac{x^2}{4x} + \frac{1}{4x^3}\right) \, dx<br />

    2∏ <br /> <br />
\int_1^2 \left(\frac{x^5}{12} + \frac{x}{12} + \frac{x}{4} + \frac{1}{4x^3}\right) \, dx<br />

    2∏ <br /> <br />
\int_1^2 \left(\frac{x^5}{12} + \frac{x}{12} + \frac{3x}{12} + \frac{1}{4x^3}\right) \, dx<br />

    2∏ <br /> <br />
\int_1^2 \left(\frac{x^5}{12} + \frac{4x}{12} + \frac{1}{4x^3}\right) \, dx<br />

    2∏ <br /> <br />
\int_1^2 \frac{1}{4}\left(\frac{x^5}{3} + \frac{4x}{3} + \frac{1}{x^3}\right) \, dx<br />

    ∏/2 <br /> <br />
\int_1^2 \frac{1}{4}\left(\frac{x^5}{3} + \frac{4x}{3} + \frac{1}{x^3}\right) \, dx<br />

    is that better?
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  10. #10
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    <br />
2\pi \int_1^2 \left(\frac{x^3}{6} + \frac{1}{2x}\right)\left(\frac{x^2}{2} + \frac{1}{2x^2}\right) \, dx<br />

    <br /> <br />
2\pi \int_1^2 \left(\frac{x^5}{12} + \frac{x^3}{12x^2} + \frac{x^2}{4x} + \frac{1}{4x^3}\right) \, dx<br />

    <br /> <br />
2\pi \int_1^2 \left(\frac{x^5}{12} + \frac{x}{12} + \frac{x}{4} + \frac{1}{4x^3}\right) \, dx<br />

    <br /> <br />
2\pi \int_1^2 \left(\frac{x^5}{12} + \frac{x}{3} + \frac{1}{4x^3}\right) \, dx<br />

    integrate at this point. don't mess with the algebra anymore ... too big a chance of making a bonehead error.
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  11. #11
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    ok so for the final result i was able to get

    <br />
\frac{41\pi}{16}<br />

    is that correct?
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  12. #12
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    Quote Originally Posted by hlpplz View Post
    ok so for the final result i was able to get

    <br />
\frac{41\pi}{16}<br />

    is that correct?
    my calculator says \frac{47\pi}{16}
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  13. #13
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    and for the arc length i ended up getting

    <br /> <br />
\frac{17}{12}<br />

    is that right?
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  14. #14
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    you're done.
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