How to use parametric equations for line integral?

Question

The question is to find the work done in moving a particle in a force field:

$$\overrightarrow F = 3x^2i+(2xy-y)j+3k$$

along the straight line from (0, 0, 0) to (2, 1, 3)

So, work done $=\int \overrightarrow F \cdot d\overrightarrow R = \int 3x^2dx+(2xy-y)dy+3dz $

I can find the values of $x, y, z$ from the equation of straight line:

$$\frac {x-0}{(2-0)} = \frac {y-0}{(1-0)} = \frac {z-0}{(3-0)}$$

$$\frac {x}{2} = \frac {y}{1} = \frac {z}{3}$$

Now, how do I find the limits? I was told to use parametric equations but I have no idea how to use them.

Is there a straight-forward way to solve this by picking up integral limits for $x, y,$ or $z$ for example?

Answer 1

You can set $y=t$, so that $x=2t$ and $z=3t$. Substitute these into your integral, together with $dy=dt$, $dx=2dt$ and $dz=3dt$. Integrate over $0\le t\le1$.

Answer 2

Well, first thing we need to do is parameterize the line segment.

Recall that the formula is:

$$\overrightarrow r(t) = (1-t)<0, 0, 0> + t<2, 1, 3> = <2t, t, 3t>$$

Then we need the vector field:

$$\overrightarrow F\left(\overrightarrow r(t)\right) = 3x^2i+(2xy-y)j+3k= 3(2t)^2i+(2(2t)(t)-(t))j+3k$$ $$= 12t^2i + (4t^2-t)j + 3k$$

We need to take the derivative of $\overrightarrow r(t)$:

$$\overrightarrow r'(t) = <2, 1, 3>$$

Now we need the dot product:

$$\overrightarrow F\left(\overrightarrow r(t)\right) \cdot \overrightarrow r'(t) = 24t^2 + (4t^2-t) + 9 = 28t^2 - t + 9$$

Finally, we integrate from $0 \le t \le 1$. Note that this is our bounds for the parameterization for the line segment:

$$\int_{0}^{1} (28t^2 - t + 9)dt = \left[\frac{28}{3}t^3 - \frac{1}{2}t^2 + 9t\right]_{0}^{1} = \frac{28}{3} - \frac{1}{2} + 9 = \frac{107}{6}$$

That was a lot to take in. Comment if you have questions.

Answer 3

In your case, the line integral along a path $ C$ in a vector field $ \mathbf{F}$ is defined as:

$$ \int_C \mathbf{F}(\mathbf{r}) d\mathbf{r}= \int_a^b \mathbf{F}(\mathbf{r}(t)) \mathbf{r'}(t)dt $$

A parametric equation of a path is an expression that describes the path using a parameter ( $ t$ in the equation above); in your case it can be expressed as the function: $$ \mathbf{r}: [0,1] \rightarrow \mathbb{R}^3 $$ $$ \mathbf{r}(t)=(2t,t, 3t) $$

By differentiating $ \mathbf{r}(t) $ and doing the dot product with $\mathbf{F}$ you will obtain a simple integral in one variable.