I was trying to solve this integral here but I am having trouble understanding if what I am doing is correct or not.
So the question says:
"Let $F(x,y) = x^2-y^2$. Evaluate:
a) $\int^{(2,8)}_{(0,0)} (\text{grad}\ F \cdot d\bf{s})$ along the curve $y=x^3$
b) $\oint \frac{\partial F}{\partial n}ds$ around the circle $x^2+y^2 = 1$. Here, $\frac{\partial F}{\partial n}$ is the directional derivative of $F$ along the outer normal and $dS$ = $|dS|$."
So I attempted part a.
I got the gradient as $\nabla F = <2x,2y>.$
If I parametrize the cubic function, I get $(x(t),y(t)) = (t,t^3) $. So then my integral goes from $t=0$ to $t=2$.
So as a result, my integral would be:
=$\int^{2}_{0} 2x\ dx - \int^{2}_{0} 2y\ dy$
=$\int^{2}_{0} 2t\ \frac{dx}{dt}dt - \int^{2}_{0} 2t^3\frac{dy}{dt}dt$
=$\int^{2}_{0} 2t(1)dt - \int^{2}_{0} 2t^3(3t^2)dt$
=$\int^{2}_{0} 2tdt - \int^{2}_{0} 6t^5 dt$
= -60
Is this right?
I am quite lost on how to do part b though. It's been a long time since I've taken multi variable calculus...
If you need to do b) by first principles
the directional derivative.
$\nabla F \cdot \frac {u}{\|u\|}\\ \nabla F = 2x, -2y\\ \frac {u}{\|u\|} = \frac {(x,y)}{\sqrt {x^2 + y^2}}\\ \nabla F \cdot \frac {u}{\|u\|} = \frac {2x^2 - 2y^2}{\sqrt{x^2+y^2}}$
The contour:
$x = \cos t, y = \cos t\\ |dS| = dt$
$\int_\gamma \nabla F \cdot \frac {u}{\|u\|} |dS| = \int_0^{2\pi} 2\cos^2 t - 2\sin^2 t \ dt = 0$