# Wire mass line integral

Wire is given with $y=\sqrt{25-x}$ and density is given with $\delta(x,y)=15-y$. Mass should be calculated using line integral.

This is my first assignment in this area and I need help with reasoning and modeling this problem as integral. Any help is appreciated.

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I've edited the answer adding some reasoning. – Ilya Oct 31 '11 at 12:32

I think that you should calculate the line integral of the first kind of the form $$\int\limits_\gamma \delta(x,y)\,\mathrm ds$$ where $\gamma$ represents your wire and $\mathrm ds$ is an infinitesimal length element. To calculate such integral we should parametrize $\gamma$. It is crucial to know what is the range of $x$: clearly $x\leq 25$ because otherwise $\sqrt{25-x}$ is undefined but what is the lower bound of $x$ (maybe it is given in your problem).
Anyway, let us consider how to solve such problem for the case when $$\gamma = \{(x,y):x\in [0,25],y = \sqrt{25-x}\}.$$ The simplest way to parametrize integral in your problem is by taking $x$ as a parameter. Then $$\mathrm ds = \sqrt{\mathrm dx^2+\mathrm dy^2} = \sqrt{1+y'^2(x)}\mathrm dx.$$ Now, $y' = -\frac{1}{2\sqrt{25-x}}$ so $y'^2 = \frac{1}{4(25-x)}$ and you line integral is $$\int\limits_\gamma \delta(x,y)\,\mathrm ds = \int\limits_{0}^{25}\delta(x,y(x))\sqrt{1-y'^2(x)}\,\mathrm dx = \int\limits_{0}^{25}\left(15-\sqrt{25-x}\right)\sqrt{1+\frac{1}{4(25-x)}}\mathrm dx$$ which value I hope you know how to find.
Now about reasoning: such integrals are used to model the mass of bodys with non-constant densities. You can think that for a part of wire with a end at the point $(x,y)$ of the infinitesimally small length $\mathrm ds$ it holds that the 'mass' of this part is $\delta(x,y)\mathrm ds$ (to be honest that is how density can be also defined). So, to find the overall mass of the wire you have to 'sum' all these infinitesimally small parts which is done by integration as usual. If $\delta$ and $\gamma$ are good enough, Riemann integration technique works well and you can think of $$\int\limits_\gamma \delta(x,y)\,\mathrm ds = \lim\limits_{\mu\to 0}\sum\limits_i\delta(x_i,y_i)\Delta s_i$$ where $\Delta s_i = \sqrt{(x_{i+1}^2-x_i^2)+(y_{i+1}^2-y_i^2)}$ is an approximation of the length of the arc from $(x_i,y_i)$ to $(x_{i+1},y_{i+1})$ and $\mu = \max_i \Delta s_i$.
Note that if the interval for $x$ is not given in your assignment, maybe the lower bound for $x$ is assumed to be $-200$ since then $\delta(x,y(x)) = 0$ and for $x<200$ the density function will be negative. – Ilya Oct 31 '11 at 12:27