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El_Arrow

  • one year ago

stuck on this surface area problem

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  1. El_Arrow
    • one year ago
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    y= (x^(2)/4)-(ln(x)/2) 1<x<2 about the y-axis

  2. El_Arrow
    • one year ago
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    @freckles @zepdrix @SithsAndGiggles @mathmath333

  3. El_Arrow
    • one year ago
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    okay so i got the set up done |dw:1443761100902:dw|

  4. El_Arrow
    • one year ago
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    when i would use u-substitution i would get |dw:1443761202970:dw|

  5. El_Arrow
    • one year ago
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    so my question is what do i do with the -(1/4)?

  6. IrishBoy123
    • one year ago
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    \(\large y= \frac{x^2}{4}-\frac{ln x}{2}\) ?? if so \(\sqrt{1 + y'^2 } = \sqrt{1 + (\frac{x}{2}-\frac{1}{2x})^2} = \sqrt{\frac{x^2}{4}+\frac{1}{4x^2} +\frac{1}{2}} =\frac{x}{2}+\frac{1}{2x}\)

  7. El_Arrow
    • one year ago
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    so thats it i dont do any u-substitution? please answer back @IrishBoy123

  8. IrishBoy123
    • one year ago
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    if you are happy with the algebra i have shown, you go straight to the integral. i think it's fine but you should check for yourself. one small correction, which **makes no difference** here, but which might matter elsewhere, is that \(\sqrt{z^2} = |z|\) so i should have said: \[\sqrt{1 + y'^2 } = \sqrt{1 + (\frac{x}{2}-\frac{1}{2x})^2} = \sqrt{\frac{x^2}{4}+\frac{1}{4x^2} +\frac{1}{2}} =\sqrt{\left( \frac{x}{2}+\frac{1}{2x} \right)^2} = \left| \frac{x}{2}+\frac{1}{2x} \right|\] and here: \(\large \left| \frac{x}{2}+\frac{1}{2x} \right| = \frac{x}{2}+\frac{1}{2x} \) because \(x>0\)

  9. IrishBoy123
    • one year ago
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    here that is again, this time fitting on the page... \[\sqrt{1 + y'^2 } = \sqrt{1 + (\frac{x}{2}-\frac{1}{2x})^2} = \sqrt{\frac{x^2}{4}+\frac{1}{4x^2} +\frac{1}{2}} \\ =\sqrt{\left( \frac{x}{2}+\frac{1}{2x} \right)^2} = \left| \frac{x}{2}+\frac{1}{2x} \right|\]

  10. El_Arrow
    • one year ago
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    so then what do i put for the x @IrishBoy123

  11. El_Arrow
    • one year ago
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    oh so i just multiply the x with the answer we got inside the square root

  12. IrishBoy123
    • one year ago
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    yes \[ S = 2 \pi \int\limits_{1}^{2} x \left( \frac{x}{2}+\frac{1}{2x} \right) \; dx = \dots \]

  13. El_Arrow
    • one year ago
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    alright thanks @IrishBoy123

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