
Step 1: Set up Stokes theorem as a line integral:
Stokes theorem states that $\iint_S (\nabla \times \vec{F}) \cdot \hat{n}\, dS = \oint_C \vec{F} \cdot d\vec{r}$, where C is the closed boundary of S traversed in the direction consistent with the chosen normal. From the figure, C is the path T(0,0) to U(4,0) to V(4,4) to W(0,4) back to T(0,0), all of which lies in the plane z = 0.
Step 2: Simplify F on the boundary:
Since the entire path lies at z = 0, substitute z = 0 into F. This gives $\vec{F} = (y + 2)\hat{i} + 8\hat{j} + 0\hat{k}$. Also, since the path stays in the XY plane, $d\vec{r} = dx\,\hat{i} + dy\,\hat{j}$, so $\vec{F} \cdot d\vec{r} = (y+2)\,dx + 8\,dy$.
Step 3: Integrate along T to U:
On this segment y = 0 and dy = 0, with x going from 0 to 4. The integrand is $(0+2)\,dx = 2\,dx$, so the integral is $2 \times (4 - 0) = 8$.
Step 4: Integrate along U to V:
On this segment x = 4 and dx = 0, with y going from 0 to 4. The integrand is $8\,dy$, so the integral is $8 \times (4 - 0) = 32$.
Step 5: Integrate along V to W:
On this segment y = 4 and dy = 0, with x going from 4 to 0. The integrand is $(4+2)\,dx = 6\,dx$, so the integral is $6 \times (0 - 4) = -24$.
Step 6: Integrate along W to T:
On this segment x = 0 and dx = 0, with y going from 4 to 0. The integrand is $8\,dy$, so the integral is $8 \times (0 - 4) = -32$.
Step 7: Add the four segments and take the absolute value:
Adding all four parts gives $8 + 32 - 24 - 32 = -16$. This matches the surface integral found from the curl, confirming the calculation. The absolute value requested by the question is 16.
Final Answer:
\[ \boxed{16} \]
Let \( R \) be the planar region bounded by the lines \( x = 0 \), \( y = 0 \) and the curve \( x^2 + y^2 = 4 \) in the first quadrant. Let \( C \) be the boundary of \( R \), oriented counter clockwise. Then, the value of:
\[ \oint_C x(1 - y) \, dx + (x^2 - y^2) \, dy \] is equal to: