# Vector Calc-Line Integral

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

$$\int_{0}^{1} 2xy+6y dt$$
$$\int_{0}^{1} 2(2+6t)(4+2t)+6(2+6t) dt = 82$$

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

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

82+184 = 266

However this is incorrect :(

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

$$\int_{0}^{1} 2xy+6y dt$$
I think the integrand should be $$\rho ds=\rho \frac{ds}{dt}dt$$
where ds is a piece of length from your curve. So if you parametrized the line by $$\vec r(t)$$, then $$ds=|\frac{d\vec r}{dt}|dt$$

Are you saying I need to do

$$\int_{0}^{1} (2xy+6x)* \sqrt{(2y+6)^2+(2x)^2}$$

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

$$\int_{0}^{1} (2(2+6t)(4+2t)+6(2+6t)) * \sqrt{(2(4+2t)+6)^2+(2(2+6t))^2}$$

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

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

Then add the results of the two integrals?

However this is incorrect :(

Galileo said:
I think the integrand should be $$\rho ds=\rho \frac{ds}{dt}dt$$
where ds is a piece of length from your curve. So if you parametrized the line by $$\vec r(t)$$, then $$ds=|\frac{d\vec r}{dt}|dt$$

Any ideas... I was not quite sure what you meanat by this post...

Ahhh I figured it out... I was doing the wrong dervatives....