Line Integrals

**s7b** · March 11th 2009, 11:22 AM

I don't understand how to go about solving a problem of this form;

Integrate f(x,y,z)=x+ sqrt(y) - z^2 over the path C from (0,0,0) to (1,1,1) given by C=C1nC2uC3, where

C1: r(t)=tk , 0<t<1
C2: r(t)=tj + k , 0<t<1
C3: r(t)=ti + j + k , 0<t<1

Anyone have any suggestions??

**Scott H** · March 11th 2009, 04:31 PM

If I understand correctly, the line integral for scalar functions is defined as

$\int_C\,f(\textbf r(t))\,ds=\int_a^b\,f(\textbf r(t))s'(t)\,dt.$

As

$ \begin{aligned} s(t)&=\int_a^t\,ds\\ &=\int_a^t\,\sqrt{dx^2+dy^2+dz^2}\\ &=\int_a^t\,\sqrt{\left(\frac{dx}{dt}dt\right)^2+\ left(\frac{dy}{dt}dt\right)^2+\left(\frac{dz}{dt}d t\right)^2}\\ &=\int_a^t\,\sqrt{\left(\frac{dx}{dt}\right)^2+\le ft(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\righ t)^2}\,dt, \end{aligned} $

we may conclude that

$s'(t)=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\fr ac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}.$

Now, these derivatives on $C_1$ , $C_2$ , and $C_3$ are as follows:

$ \begin{array}{cccc} C_1: & x'=0 & y'=0 & z'=1 \\ C_2: & x'=0 & y'=1 & z'=0 \\ C_3: & x'=1 & y'=0 & z'=0 \end{array} $

In all cases, therefore, we end up with $s'(t)=1$ . Our task is now to find

$\int_{C_1}\,f(\textbf r(t))\,dt+\int_{C_2}\,f(\textbf r(t))\,dt+\int_{C_3}\,f(\textbf r(t))\,dt.$

The function $\textbf r(t)$ is different for each $C$ :

$ \begin{aligned} C_1:\,f(\textbf r(t))&=0+\sqrt{0}-t^2=-t^2 \\ C_2:\,f(\textbf r(t))&=0+\sqrt{t}-1^2=\sqrt{t}-1 \\ C_3:\,f(\textbf r(t))&=t+\sqrt{1}-1^2=t. \end{aligned} $

**s7b** · March 11th 2009, 08:01 PM

Ok, this makes sense.

Is there a different approach when faced with a question like this;

Integrate f over the given curve: f(x,y) = (x+y^2)/(sqrt(1+x^2)) , C: y=(x^2)/2 from (1,1/2) to (0,0)

..Or is the same method applied.

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