Stokes Theorem for a Circle In the Plane x + y + z = 5
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.
1 Answer
$\begingroup$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.
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