Question

need help with a line integral ( i think). The problem is as follows:

find the mass of the wire AB ( picture below ) using the density function f(x,y)=x

Answer 1

Do you know how to parameterize a line segment?

Answer 2

that would be the issue, mainly.

Answer 3

Well there are two ways to do it. One would be to find the equation of the line, another would be to set up a vector equation.

The equation of the line connecting (1,0) and (3,1) is \(y=\dfrac{1}{3}x-\frac{1}{3}\). So you could use the following parameterization:
\[C:=x=t,~y=\frac{1}{3}t-3,~1\le t\le3\\
\text{which gives you}\\
\frac{dx}{dt}=1,~\frac{dy}{dt}=\frac{1}{3}\]
So the integral would be
\[M=\int_Cf(x,y)~ds=\int_1^3t\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt\]

Answer 4

Disregard that vector setup I mentioned. I don't think you have the right conditions to do it.

Answer 5

the line equation i got there was \[y=\frac{ 1 }{2 } x - \frac{ 1 }{2 } \] ( used the formulae ( x2-x1)(y-y1)=(x-x1)(y2-y1). regardless of that, i still don't understand why after x=t y=1/3t-3 and goes between 1 and 3.

Answer 6

That's what I get for trying to do simple arithmetic in my head -_-

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
The equation of the line connecting (1,0) and (3,1) is \(\color{red}{y=\dfrac{1}{2}x-\dfrac{1}{2}}\). So you could use the following parameterization:
\[C:=x=t,~\color{red}{y=\frac{1}{2}t-\frac{1}{2}},~1\le t\le3\\
\text{which gives you}\\
\frac{dx}{dt}=1,~\color{red}{\frac{dy}{dt}=\frac{1}{2}}\]
So the integral would be
\[M=\int_Cf(x,y)~ds=\int_1^3t\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt\]
\(\color{blue}{\text{End of Quote}}\)

\(t\) is a dummy variable. You could continue to use \(x\); I use \(t\) out of habit, as these line integral questions usually provide more complicated functions.

Suppose we just use \(x\). What would the lower/upper limits of this variable be? The first point is (1,0), whose x-coordinate is 1. The second is (3,1), with x=3. Does that make sense?

Next, you differentiate \(x\) and \(y\) with respect to the dummy variable. In my work, it's t, so you find \(dx/dt\) and \(dy/dt\), then plug them into the \(ds\) formula.

Answer 7

And finally, in the integral, since \(f(x,y)=x\), you would simply substitute and say \(f(x,y)=t\).

Answer 8

so, if instead of x we would have done the parametrization by y, with y=t and x=2y+1, t would have ranged from 0 to 1? Of course, the solution would have been different, that much I understand, it's just the parameter that I want to make sure of.

Answer 9

Actually, the solution would be the same. Just a different but equivalent integral.

Answer 10

That's what I wanted to say, sorry. same result, different way to get to it.

Answer 11

Yes, that's it.

Answer 12

And here's a check with Wolfram:
http://www.wolframalpha.com/input/?i=Integrate%5Bt*Sqrt%5B5%2F4%5D%2C%7Bt%2C1%2C3%7D%5D%2C+Integrate%5B%282t%2B1%29*Sqrt%5B5%5D%2C%7Bt%2C0%2C1%7D%5D

Answer 13

Thank you very much for the explanation. I don't know exactly why I got scared of that t there and kept looking for a complicated solution for it, but now I got it. Thank you again.

Answer 14

You're welcome!