stuck on this surface area problem

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stuck on this surface area problem

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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y= (x^(2)/4)-(ln(x)/2) 1
okay so i got the set up done |dw:1443761100902:dw|

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when i would use u-substitution i would get |dw:1443761202970:dw|
so my question is what do i do with the -(1/4)?
\(\large y= \frac{x^2}{4}-\frac{ln x}{2}\) ?? if so \(\sqrt{1 + y'^2 } = \sqrt{1 + (\frac{x}{2}-\frac{1}{2x})^2} = \sqrt{\frac{x^2}{4}+\frac{1}{4x^2} +\frac{1}{2}} =\frac{x}{2}+\frac{1}{2x}\)
so thats it i dont do any u-substitution? please answer back @IrishBoy123
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 \(x>0\)
here that is again, this time fitting on the page... \[\sqrt{1 + y'^2 } = \sqrt{1 + (\frac{x}{2}-\frac{1}{2x})^2} = \sqrt{\frac{x^2}{4}+\frac{1}{4x^2} +\frac{1}{2}} \\ =\sqrt{\left( \frac{x}{2}+\frac{1}{2x} \right)^2} = \left| \frac{x}{2}+\frac{1}{2x} \right|\]
so then what do i put for the x @IrishBoy123
oh so i just multiply the x with the answer we got inside the square root
yes \[ S = 2 \pi \int\limits_{1}^{2} x \left( \frac{x}{2}+\frac{1}{2x} \right) \; dx = \dots \]
alright thanks @IrishBoy123

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