Given $F=(3x^2+6y)\mathbf{\hat{x}}-14yz\mathbf{\hat{y}}+20xz^2\mathbf{\hat{z}}$, I am trying to evaluate the line integral $\int_C{A}\cdot d\mathbf{r}$ from $(0,0,0)$ to $(1,1,1)$. C is given as three different cases, where
- $x=t, y=t^2, z=t^3$
- straight line from $(0,0,0)$ to $(1,1,1)$
- straight lines from $(0,0,0)$ to $(1,0,0)$, then $(1,1,0)$ to $(1,1,1)$
Can anyone help me start this? I'm having a difficult time remembering exactly how this is done.
For the first integral, the path is outlined clearly: ${\bf r}=\langle t,t^2,t^3\rangle$ (excuse the sudden change in notation, it's more convenient to type). $$\int_CA\cdot d{\bf{r}}=\int_a^b \langle 9t^2,-14t^5,20t^7\rangle\cdot d\langle t,t^2,t^3\rangle=\int_a^b \langle 9t^2,-14t^5,20t^7\rangle\cdot \langle 1,2t,3t^2\rangle\,dt=\cdots$$ where $[a,b]$ is the domain you seem to have left out.
Based on your comment, you seem to have a handle on the second question. With your third question, it's unclear whether the path is continuous (i.e. from $(0,0,0)$ to $(1,0,0)$ to $(1,1,0)$ to $(1,1,1)$) or if there are two disconnected paths. Assuming the former, the work is similar to what you did for (2): You have to parameterize the four distinct paths, then integrate along them accordingly.