Surface area of x = ln(y) - y^2/8 from 1 to e

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  • #1
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Homework Statement


Find the area of the surface obtained from rotating the curve x = ln(y) - y^2/8 on [1,e] about the y-axis.


Homework Equations


SA = [itex]\int 2\pi*(f(x))*\sqrt{1+[f'(x)]^2}dx[/itex]


The Attempt at a Solution


SA = [itex]\int 2\pi*(ln(y) - \frac{y^2}{8}))*\sqrt{1+\frac{9}{16*y^2}}dy[/itex] from 1 to e

I don't know how to solve this integral, which makes me thing I set it up wron.g
 

Answers and Replies

  • #2
Dick
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Homework Statement


Find the area of the surface obtained from rotating the curve x = ln(y) - y^2/8 on [1,e] about the y-axis.


Homework Equations


SA = [itex]\int 2\pi*(f(x))*\sqrt{1+[f'(x)]^2}dx[/itex]


The Attempt at a Solution


SA = [itex]\int 2\pi*(ln(y) - \frac{y^2}{8}))*\sqrt{1+\frac{9}{16*y^2}}dy[/itex] from 1 to e

I don't know how to solve this integral, which makes me thing I set it up wron.g

I don't think you did the sqrt(1+f'(y)^2) part right. Can you show how you worked that out?
 
  • #3
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That was the equation that my prof. gave me, lol
 
  • #4
Dick
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That was the equation that my prof. gave me, lol

I'm saying I don't think you simplified 1+f'(y)^2 correctly when f(y)=ln(y)-y^2/8. You should get something inside the square root that becomes a perfect square. I'm asking how you got that part to be 1+9/(16y^2)? I don't think that's right.
 
  • #5
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oh ok. I took the derivative of [itex]x = lny - \frac{y^2}{8}[/itex] and got

[itex]x = \frac{1}{y} - \frac{1}{4y}[/itex] and squared it to get

[itex]\frac{1}{y^2} - \frac{1}{2y^2} + \frac{1}{16y^2}[/itex] which is

[itex]\frac{16}{16y^2} - \frac{8}{16y^2} + \frac{1}{16y^2}[/itex] which is

[itex]\frac{9}{16y^2}[/itex]
 
  • #6
Dick
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oh ok. I took the derivative of [itex]x = lny - \frac{y^2}{8}[/itex] and got

[itex]x = \frac{1}{y} - \frac{1}{4y}[/itex] and squared it to get

[itex]\frac{1}{y^2} - \frac{1}{2y^2} + \frac{1}{16y^2}[/itex] which is

[itex]\frac{16}{16y^2} - \frac{8}{16y^2} + \frac{1}{16y^2}[/itex] which is

[itex]\frac{9}{16y^2}[/itex]

The derivative of that is [itex]\frac{1}{y}-\frac{y}{4}[/itex]. That's where you are starting to go wrong.
 

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