Question

Consider a vector function \( \vec{F} = (y-z+2)\hat{i} + (yz+8)\hat{j} - xz\hat{k} \), on the surface (S) in the XY plane at \( z = 0 \) (as shown in the figure).
Considering Stokes theorem for space, the ABSOLUTE value of the surface integral \[ \iint_S (\nabla \times \vec{F}) \cdot \hat{n}\, dS \] is __________. (Answer in integer)

(Consider \( \hat{i}, \hat{j}, \hat{k} \), and \( \hat{n} \) as unit vectors)

Hint:

Compute curl F, dot it with the unit normal on the flat square, and integrate over the 4 by 4 region, or equivalently evaluate the line integral around T to U to V to W to T.

Answer 1

Correct Answer: 16

Step 1: Identify the surface and read the figure: 
The figure shows the surface S as the flat square lying in the XY plane at z = 0, with corners T(0,0), U(4,0), V(4,4) and W(0,4). The boundary curve C of this surface is the path T to U to V to W to T, which traces the square in the counterclockwise direction when viewed from the positive z axis, so the outward unit normal on S is \( \hat{n} = \hat{k} \).

Step 2: Compute the curl of F: 
Given $\vec{F} = (y - z + 2)\hat{i} + (yz + 8)\hat{j} - xz\hat{k}$, the curl is found from $\nabla \times \vec{F} = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right)\hat{i} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right)\hat{j} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\hat{k}$. Working out each term: $\frac{\partial F_z}{\partial y} = 0$ and $\frac{\partial F_y}{\partial z} = y$, so the i component is $0 - y = -y$. Next $\frac{\partial F_x}{\partial z} = -1$ and $\frac{\partial F_z}{\partial x} = -z$, so the j component is $-1 - (-z) = z - 1$. Finally $\frac{\partial F_y}{\partial x} = 0$ and $\frac{\partial F_x}{\partial y} = 1$, so the k component is $0 - 1 = -1$. This gives $\nabla \times \vec{F} = -y\hat{i} + (z-1)\hat{j} - \hat{k}$.

Step 3: Dot the curl with the unit normal: 
On the surface S, the normal is $\hat{n} = \hat{k}$, so only the k component of the curl survives the dot product. This gives $(\nabla \times \vec{F}) \cdot \hat{n} = -1$. This value is a constant, it does not depend on x, y or z, so it does not matter that z is zero everywhere on S, the k component of the curl is already exactly minus one everywhere.

Step 4: Integrate over the surface: 
The surface integral becomes $\iint_S (\nabla \times \vec{F}) \cdot \hat{n}\, dS = \iint_S (-1)\, dA = -\text{Area}(S)$. The region S is a square with side length 4, running from x = 0 to x = 4 and y = 0 to y = 4, so its area is $4 \times 4 = 16$ square units. Therefore the surface integral equals $-16$.

Step 5: Take the absolute value: 
The question asks for the absolute value of the surface integral, so the negative sign is dropped, giving 16.

Final Answer: 
\[ \boxed{16} \]