Find $\int_C \vec F \cdot d \vec r $ where $C$ is a circle of radius $2$ in the plane $x + y + z = 5$, centered at $(4,4,-3)$ and oriented clockwise when viewed from the origin, if $\vec F = 4y\vec i – 5\vec j + 2(y-x)\vec k$.
The chapter is on Stoke's theorem, so I'm using that method to solve it.
$$\int_C \vec F \cdot d \vec r = \int \int_S curl \vec F \cdot d\vec S$$
$$ curl \vec F = <2, 2, -9>$$
$$ d\vec S = \frac{\nabla g}{\lvert\nabla g \cdot \vec k \rvert} = \frac {<1,1,1>}{1} $$
So the integral is just $ \int \int_D -5 dA $. At first, for $dA$ i just found the area of the circle, which gave me an answer of $-20 \pi$, which was wrong. Then i switched the sign in case i messed up the orientation and it was also wrong.
I then realized that I needed to project the circle onto the xy plane and take the area of the resultant ellipse, but I'm not sure how to do this.
Best Answer
Don't project the circle in the tilted plane. Just figure out $\text{curl}\,\vec F \cdot \vec n$, where $\vec n$ is the unit normal (appropriately chosen to give the correct orientation). That will be a constant, so the answer will be that constant times the area of the circle.