Vector Calc-Line Integral

  • #1
1,015
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A wire lies along the piecewise linear curve extending from the point (2,4) to the point (8,6) to the point (8,11). If the density of the wire is given by p(x,y)=2xy+6x, use a line integral to find the mass of the wire.

Here is what I have tried:

C1: (1-t)<2,4>+t<8,6>
C1: x=2+6t y=4+2t

[tex]\int_{0}^{1} 2xy+6y dt [/tex]
[tex]\int_{0}^{1} 2(2+6t)(4+2t)+6(2+6t) dt = 82[/tex]

C2: (1-t)<8,6> + t<8,11>
C2: x=8 y=6+5t

[tex]\int_{0}^{1} 2xy+6y dt [/tex]
[tex]\int_{0}^{1} 2(8)(6+5t)+6(8) dt = 184[/tex]

82+184 = 266

However this is incorrect :(
 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
1,991
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Tom McCurdy said:
C1: (1-t)<2,4>+t<8,6>
C1: x=2+6t y=4+2t

[tex]\int_{0}^{1} 2xy+6y dt [/tex]
I think the integrand should be [tex]\rho ds=\rho \frac{ds}{dt}dt[/tex]
where ds is a piece of length from your curve. So if you parametrized the line by [tex]\vec r(t)[/tex], then [tex]ds=|\frac{d\vec r}{dt}|dt[/tex]
 
  • #3
1,015
1
Are you saying I need to do

[tex] \int_{0}^{1} (2xy+6x)* \sqrt{(2y+6)^2+(2x)^2} [/tex]

C1: (1-t)<2,4>+t<8,6>
C1: x=(2+6t) y=(4+2t)


[tex]\int_{0}^{1} (2(2+6t)(4+2t)+6(2+6t)) * \sqrt{(2(4+2t)+6)^2+(2(2+6t))^2} [/tex]

C2: (1-t)<8,6> + t<8,11>
C2: x=(8) y=(6+5t)

[tex]\int_{0}^{1} 2xy+6y dt [/tex]
[tex]\int_{0}^{1} (2(8)(6+5t)+6(8)) * \sqrt{(2(6+5t)+6)^2+(2(8))^2} = [/tex]

Then add the results of the two integrals?

However this is incorrect :(
 
  • #4
1,015
1
Galileo said:
I think the integrand should be [tex]\rho ds=\rho \frac{ds}{dt}dt[/tex]
where ds is a piece of length from your curve. So if you parametrized the line by [tex]\vec r(t)[/tex], then [tex]ds=|\frac{d\vec r}{dt}|dt[/tex]

Any ideas... I was not quite sure what you meanat by this post...
 
  • #5
1,015
1
Ahhh I figured it out... I was doing the wrong dervatives....
 

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