## El_Arrow one year ago stuck on this surface area problem

1. El_Arrow

y= (x^(2)/4)-(ln(x)/2) 1<x<2 about the y-axis

2. El_Arrow

@freckles @zepdrix @SithsAndGiggles @mathmath333

3. El_Arrow

okay so i got the set up done |dw:1443761100902:dw|

4. El_Arrow

when i would use u-substitution i would get |dw:1443761202970:dw|

5. El_Arrow

so my question is what do i do with the -(1/4)?

6. IrishBoy123

$$\large y= \frac{x^2}{4}-\frac{ln x}{2}$$ ?? 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

so thats it i dont do any u-substitution? please answer back @IrishBoy123

8. 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 $$x>0$$

9. IrishBoy123

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

so then what do i put for the x @IrishBoy123

11. El_Arrow

oh so i just multiply the x with the answer we got inside the square root

12. IrishBoy123

yes $S = 2 \pi \int\limits_{1}^{2} x \left( \frac{x}{2}+\frac{1}{2x} \right) \; dx = \dots$

13. El_Arrow

alright thanks @IrishBoy123