Given the vector field $\mathbf{E}=\mathbf{a_x}y+\mathbf{a_y}x$, evaluate $\int\mathbf{E} \cdot \text{d}l$ from $P_3(3,4,-1)$ to $P_4(4,-3,-1)$ by converting both $\mathbf{E}$ and $P_3$ and $P_4$ into cylindrical coordinates: –
My attempt
Converting $\mathbf{E}$ to cylindrical coordinates should be: $\mathbf{E} = \mathbf{a_r}r\sin(\phi) + \mathbf{a_\phi}r\cos(\phi)$.
Converting the points to cylindrical should be using $r=\sqrt{x^2+y^2}$ and $\phi = \arctan(\frac{y}{x})$:
$P_3(5, 53.13^\circ, -1) \: \: $ and $\: \: P_4(5, -36.9^\circ, -1)$
We calculate the dot product: $\mathbf{E} \cdot \text{d}l = r\sin(\phi)\text{d}r + r^2\cos(\phi)\text{d}\phi$
Finally we integrate:
$$\int_{53.13^\circ}^{-36.9^\circ}\int_5^5r\sin(\phi)\text{d}r+r^2\cos(\phi)\text{d}\phi=0 $$
And this is where I'm stuck. It can'
t be right that the result is $0$. It's that $r$ going from $5$ to $5$ that makes it all go to zero but what should I do instead? Btw, is my conversion to cylindrical coordinates correct?
Best Answer
That's not correct conversion. We have,
$\hat x = \cos\phi ~\hat r - \sin\phi ~ \hat \phi$
$\hat y = \sin\phi ~\hat r + \cos\phi ~ \hat \phi$
So in cylindrical coordinates, $\mathbf {E} = y ~\hat x + x ~\hat y$ will be written as,
$\mathbf{E} = r \sin2\phi ~ \hat r + r \cos 2\phi ~\hat \phi$
Points $P_3$ and $P_4$ are $(5, \arctan (4/3), -1)$ and $(5, \arctan (-3/4), -1)$ respectively.
So it is clear that both points are on curve $C$ which is a circle centered on z-axis at $z = -1$, with radius $5$. Also note that the vector field is conservative so using the scalar function $f(r, \phi)~$ whose gradient is $\mathbf{E}$, the line integral is simply $f(P_4) - f(P_3)$.
If you want to evaluate the line integral, it is easier to do so along curve $C$ from $P_3$ to $P_4$.
$ \displaystyle \int_C \mathbf{E} \cdot dl = 5 \int_{\arctan(4/3)}^{\arctan(-3/4)} \mathbf{E} \cdot \hat \phi ~ d\phi$
$ \displaystyle = 25 \int_{\arctan(4/3)}^{\arctan(-3/4)} \cos (2\phi) ~ d\phi = -24$