Evaluate the surface integral $\int \int (curl F) \cdot n dA $ directly for the given F and S.
$$F = [z^2, -x^2, 0] $$ and S is the rectangle with verticies (0,0,0), (1,0,0), (0,4,4), (1,4,4)
So the formula in Stokes's theorem to find the line integral involves finding the curl of F and n which is the normal unit vector... but how do we find n here?
What is S for here? Does it help us find n?
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$\begingroup$As the orientation of the rectangle is not given, here I will assume clockwise orientation (as seen from top). We can first find the plane on which the rectangle lies.
We can get directional vectors of two lines from the vertices $A(0,0,0), B (1,0,0), C (1,4,4), D (0,4,4)$
$\vec{AD} = (0-0,4-0,4-0) = (0,4,4), \vec{DC} = (1-0,4-4,4-4) = (1,0,0)$
So normal vector of the plane $\vec{N} = \vec{AD} \times \vec{DC} = \begin{vmatrix} i & j & k \\ 0 & 4 & 4 \\ 1 & 0 & 0 \end {vmatrix} = \, (0,4,-4)$
And equation of the plane is $ \, z = y$.
Unit vector $n = \frac{\vec{N}}{||N||} = \frac{(0,4,-4)}{4\sqrt2} = \frac{(0,1,-1)}{\sqrt2}$, $dS = \sqrt2 dA$ (as $|n \cdot k| = \frac{1}{\sqrt2}$).
So, $\int \int (curl F) \cdot n dS = \int \int (curl F) \cdot (0,1,-1) dA$ ...(i)
$ curl F = \nabla \times F = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z^2 & -x^2 & 0\end {vmatrix} = \, (0,2z,-2x)$
Now using $z = y$ from the equation of the plane and we can integrate using (i)
$\int \int (curl F) \cdot n dS = \int_0^4 \int_0^1 (0,2y,-2x) \cdot (0,1,-1) \, dx \, dy$
$ = 2 \int_0^4 \int_0^1 (x + y) \, dx \, dy = 20$
$\endgroup$ 17 $\begingroup$Hint: $\vec{n}$ is the unit normal vector to the surface.
The (not necessarily unit) normal vector of the plane $ax+by+cz+d=0$ is $\begin{bmatrix}a\\b\\c\end{bmatrix}$. Find the plane passing through your points and then find unit normal vector and normalize it. Alternatively you can use the cross product of the vectors connecting the points together (followed by normalization).
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