Evaluate integral over path using parametrisation

Question

Evaluate the integral of $(F.dr)$ over the path using parametrisation

I know from simpler questions in class that you must find $dx/dt$, $dy/dt$ and $dz/dt$ but where do I go from there then?

Usually in our notes the equation will be something like $(x dx + y dy)$ and I can see straight away how to do the question but this one is different or maybe it's not and I'm overthinking it.

Answer 1

Integrate each of the dimensions independently to get work done like..$$W_i=\int_0^1 e^{-x}dx\ \hat{i}$$ Similarly $W_j$ and $W_k$. Now net work done will be $W=\sqrt{(W_i)^2+(W_j)^2+(W_k)^2}$

Answer 2

A line integral of a field $\vec{F}$ along a path $\vec{r}(t)$ where $t \in [a,b]$ is given by

$$\int \vec{F}\cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t))\cdot \vec{r}'(t) dt $$

The position vector is $$ \vec{r}(t) = x(t) \hat{i} + y(t)\hat{j} + z(t)\hat{k}=t \hat{i} + t\hat{j} + t\hat{k}$$

So you need to calculate $$\vec{F}(\vec{r}(t)) = e^{-x(t)}\hat{i}+e^{-y(t)}\hat{j}+e^{-z(t)}\hat{k} = e^{-t}\hat{i}+e^{-t}\hat{j}+e^{-t}\hat{k} = e^{-t}(\hat{i}+\hat{j}+\hat{k})$$

With the velocity vector

$$\vec{r}'(t) = \hat{i}+\hat{j}+\hat{k} $$

Then calculate the dot product and integrate over the bounds of t

$$ \int_0^1 (e^{-t}(\hat{i}+\hat{j}+\hat{k})) \cdot (\hat{i}+\hat{j}+\hat{k})dt = \int_0^1 3 e^{-t} dt $$

I think you should be able to calculate the rest from there

Answer 3

Hint:

As suggested in the comment: $F\cdot dr= f_xdx+f_ydy+F_zdz=e^{-x}dx+e^{-y}dy+e^{-z}dz$, so the integral is: $$ \int_\gamma e^{-x}dx+e^{-y}dy+e^{-z}dz $$ and, with the given path $\gamma$ that gives $x=y=z=t$ for $0\le t \le 1$, this becomes: $$ \int_\gamma e^{-x}dx+e^{-y}dy+e^{-z}dz=3\int_0^1e^{-t}dt $$

can you do from this?