To evaluate an integral along the curve cut from the boundary of the cube $0 \le x \le a , 0 \le y \le a , 0 \le z \le a$ by the plane $x+y+z=3a/2$

Question

Let $C$ be the curve cut from the boundary of the cube $$0 \le x \le a , 0 \le y \le a , 0 \le z \le a$$ by the plane $$x+y+z={3a\over2}$$ How to evaluate$$\int_C(y^2-z^2) dx+(z^2-x^2) dy+(x^2-y^2) dz$$ ?

I can't even figure out what $C$ is or it's parametrization , nor even the surface $S$ to apply any Stoke's theorem . Please help. Thanks in advance

Answer 1

Here's an illustration to help you out:

The cube is the black box. The plane is the triangle with vertices on the axes. The intersection of the cube and the plane (i.e. the curve $C$) is the regular hexagon.

All the intersections of the plane with the cube occur at the midpoints of the edges of the cube.

Answer 2

$$\vec F = (y^2-z^2, z^2-x^2, x^2-y^2)$$

$$ \vec\nabla \times \vec F =2( y+z, z+x, x+y) $$

the Surface $x+y+z=\frac{3a}2 $ has unit normal

$$ \hat n = \frac 1 {\sqrt 3}(1,1,1) $$

$$ \hat n \cdot ( \vec\nabla \times \vec F ) = \frac 4 {\sqrt 3} (x+y+z)$$

which has the constant value $ \hat n \cdot ( \vec\nabla \times \vec F ) = \frac {6a} {\sqrt 3} $ everywhere on the surface of the regular hexagon.

All you need to do now is to figure out the area of the hexagon and multiply your answer by $ \frac {6a} {\sqrt 3} $ .