# Math Help - Line Integrals

1. ## Line Integrals

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??

2. 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}

3. 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.