Integral for a conservative vector field

Question

I'd like some guidance on how to solve this problem.

Given vector field: $$ \vec{F} = <y \cos(xy), x \cos(xy), \frac{1}{1+z^2} >$$ and I need to compute the integral $ \int_{C} \vec{F}.d\vec{r} $ for $C$ that is the curve given by the cylinder with radius $1$ and that connects points $P_0=(1,0,1),P_1= (0,1,0)$.

What I did was parametrize $r(t) :x = r\cos(t), y=r\sin(t), z=t$, with $r=1$ and defining $0 \leq t \leq 1$ and $0 \leq \theta \leq \frac{\pi}{2}$ as it's what made most sense given the points. But now I am left with a monster of a line integral when I parametrize $\vec F( r(t))$. What am I missing here? Or is this the way it should go?

Answer 1

Assuming you manage to find a "potential function" for $F$, or more precisely, a function $f: \Bbb{R^3} \to \Bbb{R}$ such that $\nabla f = \vec{F}$, then your integral depends only on the endpoints of the path (there is a famous theorem guaranteeing this.,which one?) So now, \begin{equation} \int_C \vec{F} \cdot d\vec{r} = \int_C \nabla f \cdot d\vec{r} = f(0,1,0) - f(1,0,1) \end{equation} (the last equality is true if the curve $C$ starts at $P_0$ and ends at $P_1$; otherwise there will be an overall minus sign on the RHS)

Now, all that's left is to find an appropriate $f$. My hint to you for this is to think about trigonometric functions.

Answer 2

Use the fact that $\vec F = m \vec a$, $\vec v = \int \vec a dt$, and the field is conservative. From my comment, I left off at $$\frac{m}{2} \int_{C} \frac{d}{dt}((\vec{v})^2) dt$$ Because the field is conservative, $\vec F$ can be derived from a scalar potential $V$ dependent only on the intial and final positions of the particle in the following way: $\vec F = -\nabla V$. Thus, the line integral is independent of the path, $C$, and it takes the form $\frac{m}{2}(v_2 ^2 - v_1 ^2)$, where $v_2,v_1$ are the final and initial velocities, respectively. Now take your formula for $\vec F$ and multiply by a $\frac{1}{m}$ term to each component ($\frac{1}{m} \vec F = \vec a$), and treat it like a constant. Integrate your new force vector with respect to time to get the velocity vector and plug in the final and initial values into the formula above.