I have this question as a calculus assignment and have been stucked on solving it for some time. so this is the question:
let F be a vector field: $$F(x,y,z)=((2/π)xsin πy) i + (x^2 cos πy + 2ye^{-z})j + (−y^2e^{−z} + 2x)k$$ calculate the work done by this field on the intersection of two manifolds: $$z=x^2+y^2 $$$$z=4x$$ what I have done so far: I thought the field would be conservative so the work done would be dependent only on the start and end point. but after calculating the CURL for this field I found out it is not conservative.
my second idea was to parametrize the intersection of the manifolds and then calculate the integral which was also impossible (if possible very complicated) because the after calculating F.dr you will reach a really scary integral :). I wonder if I can rewrite F as sum of two or more fields which at least some of them are conservative and then calculate the probably-easier integral for some non-conservative fields.
Your idea is correct.Let $F=F_{1}+F_{2}$ where $F_{1}=\frac{2x}{\pi}\sin(\pi y)i +(x^{2}\cos(\pi y)+2ye^{-z})\hat{j} −y^2e^{−z} \hat{k}$ and $F_{2}=2x\hat{k}$
Let $S$ is the surface of intersection of the solid paraboloid whose surface is $z=x^{2}+y^{2}$ and the plane $z=4x$ and $C$ is the curve of intersection lying on the plane $z=4x$.
Then $\text{curl}(F_{1})=0$ and hence so is it's integral over the closed curve of intersection . (Also follows from Stoke's Theorem)
Now $F_{2}=2x\hat{k}$
In cylindrical polar coordinates the curve is given by $r^{2}=4r\cos(\theta)$ .
So the parametrization of this curve is $4\cos(\theta)^{2}\hat{i}+4\cos(\theta)\sin(\theta)\hat{j}+16\cos^{2}(\theta)\hat{k}$ .
Now $$\oint 2x\hat{k} = \int_{0}^{2\pi}8\cos(\theta)\cdot\,d(16\cos^{2}(\theta))\,d\theta = 0 $$