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Intergation problem involving square root

  • Thread starter miglo
  • Start date
  • #1
97
0

Homework Statement


[tex]\int_{1}^{3}\sqrt{y^2+\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]


Homework Equations





The Attempt at a Solution


I don't know what to do with this integral, but i do feel like this can be simplified into something much simpler.
I tried rewriting it as [itex]\frac{1}{4}\int_{1}^{3}\sqrt{\frac{16y^6+8y^4+1}{y^4}}dy[/itex]
but I'm still not getting the same answer as in the back of the book

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
gb7nash
Homework Helper
805
1

Homework Statement


[tex]\int_{1}^{3}\sqrt{y^2+\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]


Homework Equations





The Attempt at a Solution


I don't know what to do with this integral, but i do feel like this can be simplified into something much simpler.
It can. Fill in the blanks.

[itex]y^2+\frac{1}{2}+\frac{1}{16y^4}[/itex] = (_ + _)2

edit: I take it back. I thought it was 1/(16y2)
 
Last edited:
  • #3
97
0
Is it [tex]\left(y^2+\frac{1}{4y^2}\right)^{2}[/tex]?
 
  • #4
33,644
5,310

Homework Statement


[tex]\int_{1}^{3}\sqrt{y^2+\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]
Is it [tex]\left(y^2+\frac{1}{4y^2}\right)^{2}[/tex]?
No, given the original problem.

Are you sure that you copied it correctly?

Could the problem have been this integral?

[tex]\int_{1}^{3}\sqrt{y^4+\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]

If that was the problem, the part inside the radical factors nicely.
 
  • #5
97
0
yeah that doesn't work.
well its an arc length problem
[tex]x=\frac{y^3}{3}+\frac{1}{4y}[/tex]
[tex]\frac{dx}{dy}=y^2-\frac{1}{4y^2}[/tex]
[tex]\int_{1}^{3}\sqrt{1+\left(y^2-\frac{1}{4y^2}\right)^{2}}dy[/tex]
[tex]\int_{1}^{3}\sqrt{1+y^2-\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]
[tex]\int_{1}^{3}\sqrt{y^2+\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]
 
  • #6
gb7nash
Homework Helper
805
1
[tex]\int_{1}^{3}\sqrt{1+\left(y^2-\frac{1}{4y^2}\right)^{2}}dy[/tex]
[tex]\int_{1}^{3}\sqrt{1+y^2-\frac{1}{2}+\frac{1}{16y^4}}dy[/tex]
The second line isn't right. Look again.
 
  • #7
97
0
Yeah I see what I'm doing wrong.
It's supposed to be [itex]y^4[/itex]
thanks mark44 and gb7nash
 

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