# Thread: Line Integral

1. ## Line Integral

If $C$ is any curve from $(0,0,0)$ to $(1,1,1)$ and $F(x,y,z)=(z+4y){i}+(2z+4x){j}+(2y+x){k}$. Compute the line integral $\int_{C}F\,dr$

2. Let be $r=(t,t,t)$ the curve. Then, $dr=(1,1,1)$. Note that the curve passes through the given points. Now, $F\cdot dr=(t+4t)+(2t+4t)+(2t+t)dt$ now, integrate

$\int_0^1 F\cdot dr$ and the problem will be done.

3. Originally Posted by pkrleads
If $C$ is any curve from $(0,0,0)$ to $(1,1,1)$ and $F(x,y,z)=(z+4y){i}+(2z+4x){j}+(2y+x){k}$. Compute the line integral $\int_{C}F\,dr$
Here's a different way to do it: The question implies that the path does not matter and a little checking of "mixed derivatives" verifies that. That means that this is an "exact differential"- there exist a function, F(x,y,z), such that $\nabla F= \frac{\partial F}{\partial x}\vec{i}+ \frac{\partial F}{\partial y}\vec{j}+ \frac{\partial F}{\partial z}\vec{k}= (z+4y)\vec{i}+ (2z+4x)\vec{j}+ (2y+x)\vec{k}$. Find that "anti-derivative" and evaluate at (1,1,1) and (0,0,0).

From $\frac{\partial F}{\partial x}= z+ 4y$ we get F(x,y,z)= xz+ 4xy+ g(y,z). Notice that the "constant of integration" may be a function of y and z since they are treated as constants in differentiation with respect to x.

Differentiating that with respect to y, $\frac{\partial F}{\partial y}= 4x+ \frac{\partial g}{\partial y}= 2z+ 4x$. The "4x" terms cancel (as they had to do- otherwise the integral would depend on the path) so now we have $\frac{\partial g(y,z)}{\partial y}= 2z$. Taking the anti-derivative of that with respect to y, g(y,z)= 2yz+ h(z) where now the "constant of integration" may depend on z. That is, we now have [tex]F(x,y,z)= xz+ 4xy+ 2yz+ h(z).

Differentiating that with respect to z, $\frac{\partial F}{\partial z}= x+ 2y+ \frac{dh}{dz}= 2y+ x$, so $\frac{dh}{dz}= 0$- it really is a constant.

F(x,y,z)= xz+ 4xy+ 2yz+ C.