# Thread: problem with line integral

1. ## problem with line integral

hi, this is my problem:

i need to find the mass of the curve (which is shown in parametric form):

$\displaystyle x = e^tcos(t), y = e^tsin(t), z = e^t$

from $\displaystyle t=0$ to $\displaystyle t=a$

and the mass density of the line is reversely proportional to

$\displaystyle x^2 + y^2 + z^2$

and also $\displaystyle \rho (1,0,1) = 1$

i am having trouble with it - is reversely proportional means

$\displaystyle \frac{1}{x^2+y^2+z^2}$,

and if so what am is supposed to do with $\displaystyle \rho$?

2. Originally Posted by vonflex1
hi, this is my problem:

i need to find the mass of the curve (which is shown in parametric form):

$\displaystyle x = e^tcos(t), y = e^tsin(t), z = e^t$

from $\displaystyle t=0$ to $\displaystyle t=a$

and the mass density of the line is reversely proportional to

$\displaystyle x^2 + y^2 + z^2$

and also $\displaystyle \rho (1,0,1) = 1$

i am having trouble with it - is reversely proportional means

$\displaystyle \frac{1}{x^2+y^2+z^2}$,

and if so what am is supposed to do with $\displaystyle \rho$?
Well, I've never seen "reversely" proportional. In fact I don't believe that "reversely" is a word in the English language! I suspect this is a mis-translation of "inversely proportional" which means just what you say: a multiple of $\displaystyle \frac{1}{x^2+ y^2+ z^2}$.

"$\displaystyle \rho(x,y,z)$" is the density function. While it is not a standard notation, since it is the only function given, I am sure that must be the case- and I suspect that previous examples in your text have used "$\displaystyle \rho$" to mean the density function. You are told two things:
1) $\displaystyle \rho= \frac{k}{x^2+ y^2+ z^2}$ and
2) $\displaystyle \rho(1, 0, 1)= \frac{k}{1^2+ 0^2+ 1^2}= 1$
You can use that to find k and so find the full density function.

Of course, the mass is just the density function integrated over the curve.

3. thanks,

i did mean inversely, it's just first time i have to translate that expression
to english.

in any case, my problem was that i didn't know about that k...